REAL ELLIPTIC CURVES
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REAL ELLIPTIC CURVES
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NORTH-HOLLAND MATHEMATICS STUDIES
54
Notas de Matematica (81) Editor: Leopoldo Nachbin Universidade Federal do Rio de Janeiro and University of Rochester
Real Elliptic Curves
N O R M A N L. ALLING University of Rochester Rochester, New York 14627, U.S.A.
NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM
NEW YORK
OXFORD
North-Holland Publishing Company, I981 All rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0 444 86233 1
Publishers: NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM NEW YORK . OXFORD Sole distributors for the U.S.A.and Canada: ELSEVIER NORTH-HOLLAND, INC. 52 VANDERBILT AVENUE, NEW YORK, N.Y. 10017
Library of Congress Cataloging in Publication Data
A l l i n g , Norman L., Real e l l i p t i c curves. (Not as d e matemgtica * 81) (North-Holland mathematics s t u d i e s ~. : 54$ I n c l u d e s index. 1. Curves, E l l i p t i c . I. Title. 11. S e r i e s : Notas d e matemgtica. (North-Holland P u b l i s h i n g Company) j 81. 111. S e r i e s : North-Holland mathematics s t u d i e s ; 54. QALN86 no. 81 [QA567] 510s [516.3'521 81-9655 I S B N 0-444-86233-1 AACR2
PRINTED IN THE NETHERLANDS
FOR KATHARINE
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PREFACE
I n P a r t I of t h i s monograph w e w i l l s k e t c h t h e 1 8 t h C e n t u r y s t u d y , by E u l e r and o t h e r s , of w h a t l a t e r became known a s e l l i p t i c I n P a r t I1 w e w i l l c o n s i d e r t h e work o n i n v e r t i n g
integrals.
c e r t a i n o f t h e s e i n t e g r a l s , by G a u s s , A b e l , and J a c o b i , t o form W e w i l l c o n s i d e r t h e work of W e i e r s t r a s s on
e l l i p t i c functions.
h i s e l l i p t i c f u n c t i o n s , t h e work o f Riemann on Riemann s u r f a c e s , and t h e work of K l e i n and o t h e r s on t h e e l l i p t i c modular f u n c t i o n .
Until. a p p r o x i m a t e l y 1840 t h e p a r a m e t e r s i n t h e s e i n t e g r a l s were real.
I n t h e l a t t e r h a l f o f t h e 1 9 t h c e n t u r y and d u r i n g m o s t o f
t h e 2 0 t h , t h e r e a l case was l a r g e l y n e g l e c t e d .
The p u r p o s e o f
t h i s monograph i s t o g i v e a v e r y t h o r o u g h t r e a t m e n t o f t h e r e a l We w i l l p r e s e n t t h e t h e o r y o f r e a l e l l i p t i c
e l l i p t i c case.
Many of t h e s e t h e o r e m s a r e new.
c u r v e s i n P a r t 111. Let
be an a l g e b r a i c f u n c t i o n f i e l d ( i n one v a r i a b l e )
E
o f g e n u s 1, o v e r t h e r e a l f i e l d face; then
Y
IR.
Let
Y
be i t s Klein sur-
i s e i t h e r a t o r u s , o r i t i s a n a n n u l u s , a Msbius
s t r i p , o r a Klein b o t t l e .
Y
w i l l be c a l l e d a
i.e.,
c u r v e w h e t h e r o r n o t i t h a s any r e a l p o i n t s : n o t t h e boundary Y
is not a torus.
aY
of
Let
Y
s
i s nonempty o r empty.
whether o r Assume t h a t
b e t h e number of components of
i t w i l l b e c a l l e d t h e s p e c i e s of
modulus t ,
elliptic
a p o s i t i v e r e a l number. vii
Y
Y.
s
aY;
a l s o has a geometric and
t
characterize
viii Y
Norman L. Ailing
up t o d i a n a l y t i c e q u i v a l e n c e .
D e f i n i n g e q u a t i o n s and t h e i r
modulus w i l l be s t u d i e d a n d compared t o a n a l y t i c d i f f e r e n t i a l s of
Y
t.
Automorphisms and
w i l l also be c o n s i d e r e d .
TABLE OF CONTENTS
Page vii
Preface Chapter 0.
Research Historical and bibliographic notes Prerequisites and exposition Indexing Acknowledgments
0.10 0.20 0.30 0.40 0.50
PART I.
ELLIPTIC INTEGRALS
Chapter 1. 1.10 1.20 1.30
Chapteh 2. 2.10 2.20 2.34 2.40 2.50
Chapter 3. 3.10 3.20 3.30 3.40
PART 11. Chapter 4. 4.10 4.20 4.30 4.40
1
Introduction
Examples of elliptic integrals
11
Some integrals associated with an ellipse The simple pendulum The lemniscate integral 21
Some addition theorems Examples of addition theorems The arcsine integral Fagnano's theorem Euler's addition theorem Other addition theorems Development of some discoveries made prior to 1827
33
Linear fractional substitutions Generalized Legendre form Some of Legendre's work Gauss's arithmetic - geometric mean ELLIPTIC FUNCTIONS 59
Inverting the integral
...
Abel's Recherches Jacobi's Fundamenta Nova Gauss's work on elliptic functions The question of priority ix
...
Norman L. Alling
X
Chapter 5. 5.10 5.20 5.30
Chapter 6. 6.10 6.20 6.30 6.40 6.50
Chapter 7. 7.10 7.20 7.30
7.40 7.50 7.60
Chapter 8. 8.10 8.20 8.30 8.40 8.50 8.60
Chapter 9. 9.10 9.20 9.30 9.40 9.50
Origins Definitions Properties of theta functions The introduction of analytic function theory
10.40 10.50
85
Early history Lattices in Q: Fields of elliptic functions Some applications of Cauchy's and Liouville's work Theta functions treated with analytic function theory Weierstrass's work on elliptic func$ions
103
+...
Introduction Weierstrass's Vorlesun en Weierstrass's t eory reordered Representation of doubly periodic functions An addition theory for 'p A relation between Weierstrass's 0 function and Riemann surfaces
135
Introduction Definitions Some properties of the Riemann sphere Some properties of @/L Surfaces of genus one The divisor class group The eiliptic modular function
167
Introduction Definition and elementary properties Reflection of J across acl D c Modular functions An inversion problem
Chapter 10. Algebraic function fields 10.10 10.20 10.30
75
Theta functions
Definitions and introduction Extensions The Riemann surface of complex algebraic function field A theorem of coequivalence The Riemann-Roch theorem
189
Table of Contents
PART 111.
REAL ELLIPTIC CURVES
Chapter 11. Real algebraic function fields and compact Klein surfaces
Chapter 12. The species and geometric moduli of a real elliptic curve
Chapter 13. Automorphisms of real elliptic curves
237
The automorphism group of Yslt The orbit subspaces of Yslt Orthogonal trajectories
Chapter 14. From species and geometric moduli to defining equations
251
Introduction Species 2 and 1 Species 0 Other quartic defining equations
Chapter 15. The divisor class group of 15.10 15.20 15.30
217
The extended modular group Species Geometric moduli Real lattices
12.10 12.20 12.30 12.40
14.10 14.20 14.30 14.40
207
Real algebraic function fields Klein surfaces Symmetric Riemann surfaces A theorem of coequivalence
11.10 11.20 11.30 11.40
13.10 13.20 13.30
xi
Introduction Calculations on Applications
YsIt
XT,
Chapter 16. Analytic differentials 16.10 16.20
273
283
Introduction Computations
Chapter 17. From defining equations to species and moduli 289 17.10 17.20 17.30 17.40 17.50
Introduction Determining the species Transformations of defining equations a. and X The invariant B
Bibliography
335
Index
341
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CHAPTER 0 INTRODUCTION
0.10
RESEARCH
In 1971 the Foundaticns of the theory
of
Klein surfaces [6], by Newcomb Greenleaf and the author, appeared.
In it a short description was given of compact Klein
surfaces of genus one, here called pp. 60-661.
elliptic curves 16,
These curves are of the following topological types:
a torus, an annulus, a Mobius strip, and a Klein bottle.
Ana-
lytic tori have, of course, been extensively studied in the classical literature.
These will be called classical elliptic
curves; the others will be called =-classical. The study of non-classical real elliptic curves is the research problem that motivated this study.
These curves are,
of course, characterized topologically by the number of boundary components they posess.
This number will be their species.
non-classical real elliptic curve
Y
of species
s
terized up to dianalytic equivale'nce [6, 951.2-1.41 real number and
t > 0
t,
its geometric modulus.
Given
s
A
is characby a positive
E
{2,1,0)
the non-classical real elliptic curve of species
and geometric modulus
t
will be denoted by
Yslt.
s
Further ,
are dianalytically equivalent if and only if Y s r t and Y S l I t l s = s' and t = t'. These results can be found in Chapter 12. The field "functions" on
E(Y
) (or E for short) of all meromorphic s,t Ys, (defined in [6, 11.31) is an algebraic
1
2
Norman L. Alling
function field (in one variable) over the reals of genus 1, whose field of constants is IR. x
and
y
such that
over IR(x).
As such it contains elements
.
E =IR(x,y
Necessarily
An algebraic equat on for
called a defining equation for
E.
is algebraic
over IR(x)
y
If
y
s =
2 or 1
Weierstrass p -function and its derivative are in E = IR(9,n') :
If
s = 0
will be
then the E.
Further,
and, of course, the following holds:
then this is not the case.
An elliptic function Q
can be defined using, for example, the Weierstrass zeta function (14.32:2), such that
E(Y
Ort
) =
IR(Q,Q') and, for which the
following holds: (2)
(Q')2
=
-(Q 2 + a 2) (Q2 + b 2 ) ,
To calculate
a
and
b
with
in terms of
a > 0
and
b > 0.
t, Jacobie's theory of
elliptic functions proves very useful.
These results can be
found in Chapter 14. Let (3)
P(x) E Ax4 + 4Bx3 + 6Cx2
be of degree E :: IR(x,y)r
n
=
where
field over IR exist unique
+
4Dx + E
E
IR[x]
3 or 4, and have distinct complex roots.
y2
=
P(x) : then
E
is an algebraic function
of genus 1, whose constant field is IR. s
E
{2,1,0) and
E(Yslt) are IR-isomorphic.
t > 0
Let
s
such that and
t
tively the species and geometriE modulus of
E
and
t
from the coefficients of
P(x).
There
and
be called respecE.
The problem
addressed and solved in Chapter 17 is the following: s
Let
compute
On developing ideas
that go back at least to Euler and Legendre one can see (53.2)
Introduction
that
y2
3
can be transformed, using linear fractional
= P(x)
F2
transformations with real coefficients, to (4)
F(2)
=
where
=
Lu I V I W (2,k) u,v and w
5 (I)
=
where
(-l)u(l-(-l)vit2)( l - ( -wlk2ii 2 ) ,
are in
and
{0,1}
k
is
(0,1]
E
Legendre's modulus. Computing the species n
let
= 3
from ( 3 ) or ( 4 ) is very easy.
s
be regarded as a real root of ( 3 ) .
m
the number of real roots of (3); then r > 0
then it turns out that
s
=
r
r/2.
Let
2 , or 0.
= 4,
Assume that
If r
be
If r
=
0.
is then necessarily either positive definite or negative
P(x)
definite.
If positive definite then s = 0 (17.20).
definite then
s = 2,
and if negative
Going back to ideas of Cayley
and Boole (c. 1 8 4 5 1 , (5) let
=
G2(P)
let
-
AE
G 3 ( P ) E ACE
and let
A(P)
4BD + 3C2 ,
+
2BCD
G32 ( P )
-
-
AD2
-
B2E
-
C3 ;
2 7 G23 ( P ) .
Then these are invariants, in the sense of classical invariant theory, of weight 4 , 6 , and 12, respectively. (6) Let
then
J(P)
=
2 G3(P)/A(P):
is
-
of course
J(P)
the action of sense.
In 5 1 7 . 5 0 we define A ( P ) > 0,
let
B
if
A ( P ) < 0,
let
B
-
1.
2
B (P) = J ( P )
an invariant of weight
0
under
i.e., an invariant, in the contemporary
GL2("):
(7) If
Clearly
-
B (P)
Let
as follows:
Po(xo)
E
lR[xo]
be of degree
4
Norman L. Alling
3 or 4 , with distinct complex roots, and let y o = P (x ) . 0 0
where
Let
so
Eo :W(xo,yo),
be the species of
Eo.
The main
theorem of Chapter 17, and perhaps of the entire monograph,is the following: Theorem 2 (§17.55). s = s
only if
and
0
F to
F o :Eo(i)
and
Eo
are 1R-isomorphic if and
@(PI = @ ( P o ) .
It has been known and
E
for perhaps a century
are C-isomorphic if and only if
always contains several subfields K E(Yslt)i for different values of
[F:K] = 2
and
that
F = K(i).
2 , 3 , or 4 , depending on
F
3
E(i)
J(P) = J ( P o ) .
that are lR-isomorphic s
and
t,
such that
(The number of such subfields is E.)
Hence, J(P)
is too course an
invariant to characterize non-classical real elliptic curves. The automorphism group length in Chapter 1 3 . Go
of
G
G
of
YsIt
is considered at
The connected component of the identity
is isomorphic to the circle group.
The orbits of
Some of these ys,t constitute a foliation of Ys,t. orbits are distinguished: for example the boundary components
Go
in
for s = 2 and 1. However, also has distinYs,t guished orbits, in spite of the fact that as a topological space
Of
it is homogeneous. the orbits of
Go
The family of orthogonal trajecteries to in
Ys,t
also gives a foliation of this
space.
0.20.
Historical a n d bibliographic notes
The history of the theory of real elliptic integrals, functions, and curves has two intervals of substantial activity: from about 1750 to the latter part of the 19'th century: and
Introduction from about 1970 to the present.
5
The results presented in Part
I11 depend heavily on the work of Euler, Legendre, Gauss, Abel, Jacobi, Cauchy, Liouville, Weierstrass, Klein and others, which apply to the real case.
For example, Abel's defining equation
was (+'I2 = ( 1
(1)
-
c 24 2) (1 + e 24 2) ,
with
c
which can be used to define a Mobius strip
and
YlIt;
e > 0, where as
Jacobi's was (2)
(sn'I2
= (1
-
2 2 2 sn ) ( 1 - k sn 1 ,
with
which can be used to define an annulus
0 < k < 1,
YzIt.
(Hence, although
equations of type (1) and type (2) are equivalent over rc are not equivalent over IR.)
they
The initial inversion of real
elliptic integrals of the first kind to form elliptic functions, by Gauss, Abel, and Jacobi, was later superceded by using theta functions, having general complex parameters, to define elliptic functions; thus the case of complex defining equations become dominant, leaving the real case to languish, even though it is in the real case in which many of the applications to physics and engineering lie.
Thus the research in this monograph is
highly dependent on the results obtained on the real case between 1750 and late in the 19'th century.
Since there seems to be no
unified treatment of this subject, which prepares the way for the real case whenever possible, the author has written one in Part I and I1 of this monograph, as an introduction to Part 111. In addition,the author has written many historical and bibliographic notes throughout this volume.
These are included
in an effort to show what the founders of the subject actually wrote.
Further, those notes could serve as a guide to possible
Norman L. Alling
6
further reading.
It should be noted however that these notes
have not been prepared on the basis of the depth of historical research required to produce a full history of this subject. For example, as the references make clear, the author depended heavily on collected works, and with few exceptions did not go back to the original published versions, much less to manuscripts, letters,
... .
He also used some secondary sources extensively:
e.g., Klein's Vorlesungen uber die Entwicklunq der Mathematick
im 19.
Jahrhundert [401.
0.30
Prerequisites
and
exposition
This volume was written to be accessible to readers whose knowledge of mathematics encompasses at least the following: 1.
Analysis:
...
calculus, elementary real variable
theory, and the contents of a standard half year course on analytic function theory (see for example the contents of Ahlfors' Complex Analysis [2], Chapters I 2.
Algebra:
-
IV).
Elementary facts about groups, rings, fields,
and vector spaces as presented for example in van der Waerden's Modern Algebra vol. 1 [61]. 3.
Topology:
Elementary facts about compact spaces, con-
nected spaces, continuous maps, and surfaces. The reader only so equipped should be able to read this monograph, but it will require effort and not a little time.
A
more highly educated reader may want to skip, or at least only scan, large sections of Parts I and 11, and concentrate on Part 111; whereas an expert in this subject may want to go directly
to Chapter 12.
With this in mind a great many interval refer-
ences have been given:
thus it should be possible to pick up
Introduction
7
this text at almost any point and be able to find out the exact meaning of the notation and terminology easily.
Although largely
expository, Parts I and I1 are sprinkled with novel approaches and may even have a few new results.
0.40
Indexing
The decimal system of indexing is used to index the sections of this volume.
(According to Whittaker and Watson, who use this
system, it goes back to Peano 169, p.11 . I
The chapters are
numbered from 0 to 17, this being Chapter 0.
The first digit
to the right of the decimal point is the index of the major divisions of the chapter in question.
The second digit to the right
of the decimal point is the index of the division of the major divisions of the chapter into minor subdivisions: thus 53.12 refers to Chapter 3, main division 1, subdivision 2.
Should we
want to refer to all of the first main division of Chapter 3 we would refer to it as 53.1. For later reference, increased clarity, and occasionally for emphasis, many expressions have been displayed. how they are numbered consider an example.
TO describe
Inside 53.12 the
first displayed expression would be referred to merely as (1). Outside 53.12 it would be referred to as (3.12:l).
0.50
Acknowledgements
Thanks are due my colleagues Professors William Eberlein, Richard Mosak, Saul Lubkin, Arnold Pizer, and others who actively participated in a seminar held during the 1977-78 academic year at the University of Rochester on elliptic functions, in which a preliminary version of part of this monograph was presented.
The
Norman L . A l l i n g
8
a u t h o r i s a l s o g r a t e f u l t o t h e s t a f f s o f t h e Rush Rhees and t h e Chester Carlson L i b r a r i e s , a t Rochester, f o r t h e i r a s s i s t a n c e ; and f o r t h e e n l i g h t e n m e n t o f t h o s e i n c h a r g e o f book a c q u i s i t i o n a t t h e U n i v e r s i t y d u r i n g i t s f i r s t c e n t u r y , f o r most o f t h e b i b l i o g r a p h i c r e s e a r c h f o r t h i s volume was done u s i n g l o c a l l y h e l d books.
Deep t h a n k s a r e d u e Mrs. S a n d i A g o s t i n e l l i , M r s .
R o b e r t a Colon and M r s . Marion L i n d who so a b l y t y p e d t h i s manuscript.
The a u t h o r w i s h e s t o t h a n k D r . Mika S e p p a l a , o f
H e l s i n k i and P i s a , whose comments on t h e e a r l y c h a p t e r s o f t h i s work were v e r y h e l p f u l .
The a u t h o r i s a l s o i n d e b t e d t o t h e s t a f f
o f t h e North-Holland P u b l i s h i n g Company who have a i d e d i n t h e p r e p a r a t i o n and p r e s e n t a t i o n o f t h i s m a n u s c r i p t .
Finally,
s p e c i a l t h a n k s a r e d u e P r o f e s s o r Leopoldo Nachbin f o r e n c o u r a g i n g t h e author t o p u b l i s h t h i s unusual mixture of r e s e a r c h , scholars h i p , and e x p o s i t i o n i n t h e N o t a s d e Mathem6tica s e r i e s .
PART I ELLIPTIC INTEGRALS
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CHAPTER 1 EXAMPLES OF ELLIPTIC INTEGRALS
Some
1.10
Let
a
i n t e g r a l s a s s o c i a t e d w i t h a n ellipse.
and
b
be positive real numbers and let
then this set is the graph of an ellipse in standard position. a 1. b
Assume that (2)
( 1 - b 2/a 2)
k
and let
'.,
then
k
is the eccentricity of (1). Clearly
k=O
if and only if
a=b:
0 0,
a' > b'. (6) Let
Let
kl
y(El1ipse (arb)) t Ellipse (a',bl). be the eccentricity of
kl = (a - b)/(a
1.11
Let
maps
-
= (1 kl)/(l + k') ,
P(0)
Clearly
€IE I R .
of
+ b)
y(El1ipse (arb));then
be the vector
0 EIR + P( 8 )
maps IR
[Or2T) injectively onto
and
0 < kl < 1.
(acose, bsin8) , onto
for all
Ellipse (arb), and
Ellipse (arb). The arc length I
Ellipse (arb) is
On letting
Cp E 0 + ~ / 2 then (1) becomes
(2) j2'(a2cos2
Cp
+ b2sin2C p ) %dCp,
which equals
0
is an elliptic integral. Let us also write out (1) in Cartesian coordinates.
On letting
x
E
at
(6) 4 a \01 f1- k2 ti 7 d t
=
It is
(5) becomes
2 2 (1-k t )dt = 4aj0 [(1-t2) (1-k2t2 1 1 4 ~
1 dt 4al . 2 1 14 0 [(l-t2) (1-k7 t
-
4ak2
1
t2dt 2 4 I 0 [(1-t2)(1-k t2)1
Examples of Elliptic Integrals
13
Following Legendre [49] X
dt j0 [ (1-t2)(1-k2t2)1%
(7)
for
X E
[-1,1],
is called an elliptic integral of the first kind, and
IX
(8)
t2dt 2 2
0 [(1-t2)(1-k t 1 1
4'
for
X E
[-1,1],
is called an elliptic integral of the second kind. Legendre's time,
k
After
became known as Legendre's modulus and
kl
(1.10:3) became known as its complementary modulus.
if
x
then (7) and (8) are improper; however they do
1
= ?
converge.)
(Note:
Elliptic integrals of the first kind play a cardinal
role in the whole theory.
Let
a
f
arc sinx;
then (7) equals
In this form it is also called an elliptic integral of the first kind. --
(10) K then
K(k)
3
(11) K'
d8 jr2(l-k 2sin2f3)'
'
is known as a complete elliptic integral
K
kind). -
Let
Let Z
K(k').
Further , 2n
2n
dB
a 0 (1-k2sin20)'
d6
=
4K -.
a
(of the
first
Norman L. Alling
14 1.12
The history of the problem of computing the arc
length of an ellipse seems to go back at least 1655, when
J. Wallis is reported, by Fricke [25, p. 1821 , to have begun considering it. worked on it.
Later various members of the Bernoulli family Most of the results and notation found in S1.11
can be found in the three volume opus by Legendre [ 4 9 ] c. 1825, on the subject.
1.20
The
s i m p l e pendulum
Consider a point mass less rigid rod of length
m
attached to one end of a weightthe other end being attached to a
R,
frictionless fulcrum that restricts the rod in such a way that it always moves in a plane
P
through the fulcrum. Assume
further that the gravitational force vector is constant, of length
g,
and that
- if
also lies in the plane
P.
it is drawn from the fulcrum
- it
Assume further that the whole appa-
ratus is in a perfect vacuum, and that Newton's Laws hold. a
be the amplitude of the pendulum and assume that
Thus we have the following diagram, where
8
E
Let
0 < a 01,
$u
cU+:
4 is called the upper half plane. (5)
=IR u
{m})
{m}.
Let
f (9)) = 4 1 .
co%Z:
Clearly the sets defined in (2), (3), and (5) are subgroups of conC.
Further,
con&
is clearly a subgroup of index 2 in
CO%C.
3.13
Given
(2) f ( z ) :h(M) ( z ) Since
1. m
M
If
f
is non-singular, f uOeC
is a pole of
is not a pole of
is an element in (3)
h
(az+ b)/(cz
f;
conC.
+ d) E
is a rational function of degree
f,
let
K(z).
let
f(u 0 ) E m .
f(m) :a/c.
Thus
If f:o
cf0, E
C
+
then
f(o)
E
It is easy to see that
is a homomorphism of
GL2(c)
into
conC, whose kernel
is C*I. (4)
Let
Clearly
Aff2(K) Aff2(K)
the affine group.
IG g)
f
E
GL2(K)1
is a subgroup of
. GL2(K).
It will be called
It is easy to see that
(5) h(Aff2 (C)) = conCm. (6)
then
Let
Tbf
r
b,
C
-
,
h(T b ) Z t b will be called translation by
b.
Clearly
Norman L. A l l i n g
36 d e t Tb = 1.
(7)
Let
h(Da) 5 d
9,
= (i
D~
for
a
E
K*.
&
w i l l be c a l l e d d i l a t i o n
a
a.
Clearly
= a.
d e t (D,)
F u r t h e r , it i s c l e a r t h a t
ITb:
(8)
b
K) u {Da: a
E
E
K*j
generate
h(S) 5 s
w i l l be c a l l e d
inversion.
that
i s o f o r d e r 2.
If
Let
s c1
:- d / c ( E K ) ;
then
Aff2(K).
-
M E GL2(K)
is in
MTaS
d e t ( S ) =1, a n d
Note t h a t
Aff2(K),
Aff2(K).
c # 0.
then
Hence
( 1 0 ) G L 2 ( K ) = Aff2(K) u (Aff2(K))S. Clearly
s
permutes
and
0
s (conCm)s = conC
Further,
0’
m
and
and maps
onto i t s e l f .
K*
t (conC ) t-l = conCa.
Hence
ow.
c1
w e have shown t h a t (11) t h e s u b g r o u p s
(conCa) a E C
z E C i s a f i x e d p o i n t of (12) cz2
+
c=O
If
and
addition, c#O f
(d - a ) z
or
i f and o n l y i f
- b = 0.
a=d,
b=O
f
are a l l c o n j u g a t e .
then
d f O and
t h e n every p o i n t of
afd,
f(z) =z+b/d. C
then ( 1 2 ) has a r o o t i n
If, in
i s f i x e d by C.
f.
If
Thus i n a l l cases
has a f i x e d p o i n t ; hence
( 1 3 ) c o n 1 = uaEcconCa. C l e a r l y ( 1 2 ) i s t h e zero polynomial i f and only i f f f l
t h e n ( 1 2 ) h a s a t most t w o r o o t s .
then
c#O
f # l
implies
(14) If
f
and h e n c e f
m
f = l .
If
I f it has t w o r o o t s
i s n o t a f i x e d p o i n t of
h a s a t most two f i x e d p o i n t s .
h a s t h r e e o r more f i x e d p o i n t s t h e n
f;
thus
Thus w e h a v e :
f=l.
37
Development of Some Discoveries Made Prior to 1827
Clearly
(15) the subgroups
conIw
=
-1 ta (conCmlo ) ta
(i.e., given
element
a
10'
conCm
acts simply transitively on
10
a'
and
(dciIia)in
conCw
=
Further, it is clear that C*
thus
E @ ;
I
(conCwla)aEc are all conjugate.
Ida: a € @ * I
(16) Clearly
a
for all
,a
in
there existsa unique
@*,
that maps
conCm
10
a').
to
ci
Combining
all of these observations we have the following: h
Theorem.
maps
GL2 ( C )
having as kernel C X I .
Given two triples
of distinct points of
C,
that
f ( a ) = a',
exactly triply transitively Given Let
M
-
E
GL2 (c)
M ' 5 1-1 'M;
then
and
on
E
SL2 (a:)
Ca',B',y')
f ~ c o n C such
i.e.,
f(y) = y':
conC
acts
C.
there exists M'
and
(a,B,y)
there exists a unique
f(B) = B ' ,
con1 ,
homomorphically onto
p
E
and
C*
h(M')
such that =
h(M)
.
.
p 2 = det (M)
Hence
(17) h(SL2 ( c ) ) = conC. Clearly
(kerh) n SL2(@)
{+I}.
2
@.
(18) PSL2(C)
and
con1
are naturally isomorphic.
co%C,
it is clear that
Further, it is evident that
(:(co%C)
n
of
Clearly
Turning now to
3.14
co%C.
is denoted by
SL2(C)/{+I}
and is called the projective special linear group
PSL2(@), rank
=
h(Aff2 (IR) )
=
h(GL2m))
co%Cm
(conZm)); thus, arguing as we did in 53.13 one sees
that (1)
h (GL2aR) )
=
co%C.
Now assume that zc + d # 0,
c
M
is in
GL2(lR).
For
z EC,
for which
Norman L. Alling
38
aczz + adz + bcz + bd 2 Icz +dl
h(M) ( z ) =
(2)
I
and hence
h(SL2@?)) c conb.
We immediately see that
6,
h(M) ( 0 ) =
+
GL2 (El)
4)
Given
M
E
i
GL;
Let
M' :p-'M;
(5)
h(GL:OR))
{z:
:
where {M
E
then
(kerh) n SL2 OR)
E
a}.
det(M) < 0
then
Let
G L 2 m ) : det(M) > 0 1
(I? there )
=
z
If
exists
M'
E
=
h(M')
=
p2 =
h(M).
.
det (M) Hence
conq. SL2 (lR)/{-tI} 5 PSL2 OR)
Let
{?I).
such that
SL2(lR) and
h(SL20R))
=
1-1 EIR
and let it
be called the projective special linear group of rank
2
Clearly
R.
(6) PSL20R)
and
conQ
is of index
ates
co%(C)/confi
con8 in
2
co%C,
and the image of
if and only if
The action of
of the action of
are naturally isomorphic.
on I R u
co%C
conC
on
C.
h(M)
gener-
det(M) < 0. {m}
is a great deal like that
Proceeding much as we did in
53.13, we obtain
acts exactly triply transitively on IR u i m 3 .
(7) co%z
Let the usual orientation of IR Ru
{m}.
Clearly
h(M)
induce an orientation on
either preserves or reverses this orien-
tation. (8) h(M) '
for all
z
( 2 )
=
det(M)/(cz +d)2 ,
for which
orientation of IR u
{m)
c z + d f 0.
Thus,
if and only if
h(M)
preserves the
det (M) > 0.
From ( 5 ) ,
Development of Some Discoveries Made Prior to 1827
39
and the remarks above, we obtain the following: (9) Each
fE
C
O
preserves the orientation of Q,
~
hence of its boundary, I R u
Each
{m}.
f
E
and
co%C
-con8
reverses the orientation of I R u (-1. Given
c1
c u+
E
(3.12:4),
let
conQa
(con&) n (conCa)
From (2) we see that ( 1 0 ) h(M) (i) = (ac + bd) + (ad - bc)i
c2 + d2 From this one easily sees that C
- mu
r
E
&.
The subgroups
M E SL2 W) R
and that
{a}),
such that
con&
co%I
acts transitively on
acts transitively on
(conQT).cES) are all conjugate. h(M)
E
~01-4~;
then
(
)
cose
Let
may be chosen in
sin0
Me.
-sin8 cose
Further,
0 EIR
-+
h(Me)
is a homomorphism of IR
since Q
onto
conQ
it is isomorphic to the circle
whose kernel is 2vZ; thus corni 1 group S (%lR/ZvZ). That is - of course
-
not at all surprising,
is conformally equivalent to the open unit disc in C.
Clearly if
f
E
conQi
Clearly the orbits of
fixes one point in I R u corni
then f = 1. are homeomorphic to S 1 ; or it
{i}.
-1 ( 1 ) M-l = (ad-bc)
If
0
Let
such that
(11) M =
is
Q.
detM= 1
d
(-c
-b a) cGL2(K).
then, of course, detM-'=
1.
We will be concerned with the equation 2 (2) Y = P ( x )
E
K[x]
,
{m},
40
Norman L. Alling
where
%
(3) Let
Let
Z
then
x
.
Then
~=P(c%+d)-~.
2
P 2 (c%+d)-4 = y2 = P ( x )
=
=P(h(M) ( % ) I ;
(c%+d)4 (P(h(M)( % ) ) I = :(%I
y2=
.
h(M) ( 2 )
h(M-I) ( x ) ;
y-y(c%+d)
Clearly (4)
is admissible (3.1).
P(x)
Lemma.
6(%)
Proof.
Let
hence
EK[%].
is admissible. 6
6(%).
be the degree of
Clearly
514.
Let
zj
(5)
Thus
h(M-l) ( p j ) ,
p j = h(M)
("p),
for each
distinct so are the four
G4
of (6)
can c# 0
in
Is.
j
Let
of course a/c
-
be zero even if
is a root of
Assume that
is non-singular).
to the finite roots of
(7) if M E Aff2(K) Now assume that
.
a # 0. then
P(x),
McAff2(K);
The finite roots o f
q2
then
(For example, if
x= 0
c=O
P(x)
then
p ( % ) is admissible.
M=S
= A-4B%+6CS2 - 4 D %
(3.13:9); 3
then
% = - l / x and
+E% 4 ,
is a factor of
P(x),
which is absurd, since
sible; thus
is always 3 or 4.
' j
' s
a#O#d
$ ( % ) ; thus
If
all the
and hence
transform, according to ( 5 ) ,
5=4.
EfO
and hence
From (4) one immediately sees that
then
If
'j"
then
:(%);
-
and
are denote the coefficient
i
Aa 4 +4Ba 3c + 6 C a 2c 2 +4Dac3 +Ec4
(since M
(8)
5
Since the four
j.
=
a2, such that
for
k=l
and
2.
k
Let
Proof.
(5) dkI b:-akckl criminant of Since
t
and 2 ;
k = l
Sk(x)
and
then
2 dk = ak(rk
4dk
is the dis-
- ri)2/4.
Sk(x) must have distinct roots
(6) dk
Let
for
for each
k.
be another indeterminant and consider
(7) J(t,x) 2 (bl
G
S1(x) - tS2(x).
-
Clearly J(t,x) = (al ta2)x2
- tb2)x + (cl- tc2) ~IR[t,xl.
+
Development of Some Discoveries Made Prior to 1 8 2 7
(8) Let
then
DJ(t) :(bl - tb2)2 - (al - ta2) (cl - tc2) cIR[t]; is the discriminant of
4DJ(t)
sidered to be a quadratic in
x.
J(t,x), where it is con-
Then
DJ(t) = (bi-a2c2)t2 +(a c +a2c1-2blb2)t+ (b:-a
(9)
43
c
1 1
1 2
2
d2t + m t + d l , where
m:a
c 1 2
=
)
a2C1 - 2 b1 b2 '
+
(10) m = a a [r r' + r r' - (rl +ri) (r2 +r')/2].
1 2
A
Let
1 1
2 2
2
DJ(t); then-of course- A
be the discriminant of
is
n
L
m -4dld2.
A routine calculation shows that
DJ(t)
Lemma
and
has two distinct non-zero real roots,
t2. Proof (of the Lemma).
is
Since the constant term of is not a root of
dll which is non-zero (6), 0
Since we have assumed that the equation
P(x) = O
has 4 real roots.
is
S1(x)
n=4,
DJ(t).
the number of real roots of
4,2, or 0. and
DJ(t)
S2(x)
Assume first that
P(x)
were chosen so that their
roots do not interlace; thus without l o s s of generality we may rl > r 1 > r > r' From (11) we see that A > 0, 1 2 2' establishing the lemma in this case. Assume now that P(x) assume that
has 2 real roots. that
S1(x)
are real.
Without loss of generality we may assume and that r and r; 2 2 2 2 2 A =ala2/r1-r21 Irl- ril > 0,
is irreducible in IR[x] Then
r' =rl; 1
hence
proving the lemma in this case.
Now assume that
P(x)
has no
-
real roots. Then r' k = rk for each k. Hence 2 2 2 A=a1a2/r1-r2I b 1 - r 2 I > 0, proving the lemma. Continuing with the proof of the Theorem, it is perfectly
44
Norman L. A l l i n g
al-t.a
= O f o r j equal t o 1 3 2 Note: since the t a r e d i f f e r e n t and a 2 # 0 , j happen f o r b o t h j. Assume f i r s t t h a t possible t h a t
( 1 2 ) al - t 1 a 2 = 0 :
(13) alb2=a2bl,
cl-t
and hence
and hence
c = O
1 2
- a2bl) 2
S1(x) = t l S 2 ( x )
t h i s cannot
showing t h a t
b 1 - t1b 2 = O .
J(tl,x) = O
then,
2.
t 1= a1/ a 2 '
that
2 0 = a D (a /a ) = (alb2 2 J 1 2
Then,
Were
i.e.,
or
x
(for a l l
(for a l l
x
in
in thus
@);
@)
(7)
P(x)
would n o t be a d m i s s i b l e , which is a b s u r d , p r o v i n g t h a t
( 1 4 ) S1(x)
- tlS2
( x ) = (c l
- tlc2)
cIR*.
t l # t 2 , by t h e Lemma, w e may u s e ( 1 2 ) t o c o n c l u d e t h a t
Since
(15) a l - t
a f 0 ; 2 2
thus
J(t2,x)
J(t2,x),
Since t h e discriminant of that there exists ( 1 6 ) S1(x)
alcIR
i s of degree
2
in
x.
i s zero w e see
4DJ(t2),
such t h a t
- t 2 S 2 ( x ) = ( a l - t 2a 2 ) ( x -
S o l v i n g ( 1 4 ) and (16) f o r
and
S1(x)
Lemma, w e see t h a t t h e r e e x i s t
A1,A2,
S2(x),
and u s i n g t h e
B1,
and
and
2.
B2
in
IR*
such t h a t
(17) Sk(x) =Ak(x-al)
2
+Bkr
for
k = l
Now assume t h a t n e i t h e r (18) al thus
- t .3 a 2
J(t.,x)
is zero;
i s a r e a l q u a d r a t i c whose d i s c r i m i n a n t i s z e r o .
7 Hence t h e r e e x i s t
(19) s1 ( x ) for
a1
- tjS2 (x) j = 1
and
and
a2
in
IR
( = J ( t,.x ) 1 = ( a l 7
2.
such t h a t
- t 7. a2 1 ( x - a 7. I 2
Development of Some Discoveries Made Prior to 1 8 2 7 Solving these equations for
S1(x)
and
S2(x)
(20) Sk(x) = A k ( ~ - ~ 1 ) 2 + B k ( ~ - i2i,2 )where in lR*,
for
Note that where each
k,
and
k = l
a1 - a 2 ,
and thus
P(x)
then
Ak
45
gives and
Bk
are
2. Sk
(x)= (Ak + Bk) (x - "1)
would have only one root in
hence would not be admissible: which is absurd.
Thus
for
I
and
@,
a1
#
a2.
Without loss of generality assume that ( 2 1 ) a1 > a2:
proving the Theorem.
3.21
Let us return to consider
(1) dx/P(x)',
:.>
If (3.20:3) of Theorem 3.20 holds let
As Euler did c. 1 7 6 6 . (2)
M
<SL2(R).
If (3.20:4) of Theorem 3 . 2 0 holds let
Set
y 2 =P(x)
and let
2
and
be defined by (3.15:3).
Using (3.15:4) and (3.15:ll) we find that -2
(4) y
=
-2
-2
(A1x + B 1 ) (A2x +B2)
By Lemma 3.15
eIR[x]
and
dx/y=dk/y.
g ( % ) is admissible. This is, essentially, what
Euler asserted [22, vol.XX, pp. 303-3041, c. 1 7 6 6 . Without loss of generality we may assume that (5)
IA1/B1l 5 [A2/B21.
Norman L. Alling
46
then
defined by (6), has determinant
Id,
is in
GLiW).
By ( 5 ) ,
ulvf and
O 0.
( a , b ) = b ( u , l ) = b ( l + x , l ) , and
6(a,b) =bx.
L y ( a , b ) = b y ( l + x , l )= b ( l + x / 2 , ( l + x ) * ) ,
and hence
(5)
If
x 0.
(3.12:4);
thus
r
5
r
As u s u a l T
2w
with
and
w2
3
r , s EIR.
20';
thus
Then
Im(7)
Let
hEeinT;
(3)
Let
U E
8a d e n o t e s t h e u p p e r h a l f p l a n e
then
v-u/2w,
h=e-'**e and
z :e
inr
,
inv
inu/2w); (=e
and
-nImv
IzI = e
Let
Z
= s,
€9.
(2)
C,
such t h a t
0.
wl-
+ si,
- {O})
d e n o t e t h e s e t of a l l i n t e g e r s .
Now l e t
Ihl=e-" 0.
He
etc. by inverting the integral, using
the addition theorem etc., but only for the special case of .r=iy. Usinq theta functions it is no more difficult to let run throuah all of
Q.
methods of Fundanenta
The author does not know if Jacobi's
Nova...
have ever been riqorously extend-
ed to cover the case of non-real
kcq: - {0,1}.
One of the dis-
advantaaes of usina theta functions to define
sn(u)
etc. is
that, a priori, it is not evident that theta functions have anything at all to do with elliptic integrals. show 136, p. 2161 that
sn(u), as defined by (l), satisfies
(2)
2 2 2 ( s n ' ( ~ ) ) ~(=1 - s n (u))(l-k sn ( u ) ) ,
(3)
k = B'//B' 2
However, one can
where
3'
Of course (2) establishes a very close connection between
sn(u)
and dz
1 (1-z2)(1-k2 z2 ) ] 4
-
One could adopt the point of view, which TQeierstrass seems to have adopted, that
sn(u)
satisfies ( 2 ) , subject to an initial condition: and
sn'(0) > 0.
that
is a meromorphic function on C namely
sn(0) = 0
Viewed in this way theta functions provide us
not only with an existence theorem, but a vast amount of information about the solution of (2) as well. Theorem.
sn(u)
periodic with periods and its poles are at
is meromorphic on C . 40
and
2w'.
2nw+ (2n'tl)w',
It is doubly
Its zeros are at for each
n
and
2n0+ 2n'w' n' E
2.
Norman L . A l l i n g
a2
t36, p . 2 1 5 1 .
5.32
(1)
One e a s i l y sees t h a t t h e f o l l o w i n g h o l d :
e , ( v ) = 1-
-
A
. ..
9
~ ~ C O S ~t T~ V ~ - C O S ~ K 2hV C O S ~ I T Vt
( v ) = 2hk s i n n v 0 2 ( v ) = 2h’cos~v
- 2 h 9 / 4 s i n 3 , ~ vt 2 h 2 5 / 4 s i n 5 ~ v- .. . + 2 h 9 / 4 c o s 3 ~ vt
2 h 2 5 / 4 c o s 5 ~ v+
A 9 O 3 ( v ) = 1 t 2hcos2rrv t 2h-cos4rrv t 2h C O S ~ I T V t
Although
ej(v)
and
. .. .
i s d e f i n e d by a n i n f i n i t e s e r i e s , it c a n
a l s o be g i v e n by a n i n f i n i t e p r o d u c t . and l e t
{2,4,...,2nI...1
.. . ,
v
g
Let
run through
run through
~ ~ , ~ , ~ , . . . , 2 n ~ 1 , . . .
Let
CZn(1-h‘);
(2)
then
9
B o ( v ) = C TI ( 1 - h v z 2 ) ( l - h ” z - 2 ) .
(3)
V
(1 - hqz-2
e 2 ( v ) =Ch’(z
t z-l)
II (1 + h q z 2 (1t hgz-2
I
,
and
9
e 3 ( v ) = C TI (1+ h ” z 2 ) (1+ h v z - 2 ) . v
Note t h a t t h e p r o d u c t s a p p e a r i n g i n ( 2 ) a n d ( 3 ) a r e v e r y c l o s e l y related to the p a r t i t i o n function (5.11:2).
Indeed, each can be
w r i t t e n i n terms o f t h e p a r t i t i o n f u n c t i o n f o r v a r i o u s v a l u e s of
x
and
z.
F r i c k e p o i n t s o u t 125, p. 2271 t h a t
(4)
P ($1 =
z3l4(!)
s
( e - I T I 4 s i n a - e -9T’4sin34
+ e-25n/4sin5+ -
. . . I,
Theta Functions
83
and (5)
(E) 4
= 2-k
O($)
(.1+ 2ecTIcos2$ + 2e-4TIcos4 $
+
... )
,
where
$ = $;/TI.
r = i,
Let (6)
then
thus, referring to (1), we see that
h = e-';
P(@)/el($/z)
and
B($)/e3($/w)
are constant; thus we have the following: Theorem.
F7here
r = i l Jacobi's theta functions are con-
stant multiples of suitably normalized versions of Gauss' "theta" functions, P,Q,p,
and
q.
Very near the end of the part of Abel's Recherches...
5.33
that appeared in volume 2 of Crelle's Journal in 1827 [l, vol.1,
p. 3471, Ahel qives the followins development of his elliptic function
6:
In [l, vol.1, SVIII, p. 3521 of his Recherches... siders the case in which $
e=c=l.
,
Abel con-
In this case Abel's function
is Gauss' sin lemn function, and Abel's constant
to Gauss' constant
-w.
Putting
(W/TT)
(sin(am/w))
product in the numerator of (1), gives Gauss' in (4.31:ll). Q(a),
(2)
w
is equal
with the
P(a) ,
as given
The product in the denumerator of (1) is Gauss'
as siven in (4.31:12).
Thus
$ (a1 = P ( a ) / Q(a)
Gauss wrote to Bessel, in Konigsberg, as follows, after
84
Norman L. Alling
the first part of Abel's gcherches
... had
appeared.
"I shall
most likely not soon prepare my investigation on the transcendental functions which I have had for many years
-
since 1798
-
because I have many other matters which must be cleared up. Herr Abel has now, as I see, anticipated me and relieved me of the burden in regard to one third of these matters, particularly since he has executed all developments with great stringency and elegance.
He followed exactly the same road which I traveled
in 1798; it is no wonder that our results are so similar.
To my
surprise this extended also to the form and even, in part, to the choice of notations, so several of his formulas appeared as if they were copied from mine.
But to avoid every misunderstand-
ing, I must observe that I cannot recall ever having communicated any of these investigations to others." 152, p. 1831.
CHAPTER 6
THE INTRODUCTION OF ANALYTIC FUNCTION THEORY
6.10
Early history.
In 1811 Gauss wrote, in a letter to Bessel, that function theory should be carried out
-
when possible
-
in the complex
plane; sketched what the integral should be; and asserted that what we know as the "Cauchy Integral Theorem" held.
He went on
to mention [27, vo1.8, pp. 90-911 "This is a very beautiful theorem whose proof (not difficult) I shall give at a suitable opportunity.
It is connected with other beautiful truths touch-
ing on expansions in sums".
Gauss waited until 1832 [27, vo1.8,
102 ff.] to publish some of his results about complex function
theory.
6.11
Augustin-Louis Cauchy (1789-1857), one of the found-
ers of analytic function theory, apparently had his Integral Theorem in mind as early as 1814, when he read a paper on the subject to the Paris Academy; however his argument in its support did not appear in print until 1825.
(See [ll, pp. 33-37] for an
English translation of some of this very interesting early paper of Cauchy.)
His Integral Formula appeared in 1841 and his theory
of residues appeared in 1826.
(See e.g.,
111, pp. 31-441 for
exact references.) Cauchy's manuscripts are easily accessible and surprisingly 85
Norman L. Alling
86
contemporary in style.
The underlying topology of the complex
plane had yet to be worked out, and the idea of uniform continuity had not yet emerged. still shaky
The existence of the integral was
by present standards of rigor.
However, the
assertions and arguments are presented in a very plausible way. Clearly it can easily be brought up to present standards of rigor by supplying a supplementary argument here and there.
6.12
Joseph Liouville (1809-1882) stated, in 1844 (see
e.g., 111, p. 32 ff], that every doubly periodic analytic function is constant.
This theorem
-
of course
-
played an important
role subsequently in the theory of elliptic functions, as we will see.
Its generalization to bounded entire functions was
called "Liouville's Theorem'' by Jordan in his Cours d'analyse [38,
p. 3081. The idea of the periodparallelogram emerged in 1847 out
of the work of Liouville and others.
(See [25, pp. 232-2331
for historical details.)
6.13
It is not the purpose of this work to trace the
development of analytic function theory from its origins, in the work of Gauss and Cauchy, to maturity at the hands of e.g., Riemann and Weierstrass.
Suffice it to say that, during the
19'th Century, it developed into the powerful and magnificent theory we know today. We will develop
in this chapter some of the consequences
of Cauchy's theory of residues and of Liouville's Theorem as applied to elliptic functions.
However, first we will make some
remarks about lattices in the complex plane, using contemporary
The Introduction of Analytic Function Theory
87
terminology and methods.
6.20
& I
Lattices
By a lattice
@,
is meant a discrete subgroup L
(1:
that is a free Abelian group of rank
the additive group of C 2:
i.e., it is a free 2-module of rank 2 .
{ w l r o 2 ) of
L=
ZWl@ZW
Given a (free) basis
then
L
will be regarded as a basis of (2)
of
L.
Clearly
2 '
let Q denote the field of all rational numbers. Theorem.
A
necessary and sufficient condition for
to be a basis of some lattice (4)
(01p021
(5)
T :w2/w1
is not in
n=m=O,
Thus if
for
L
n L
[0,1)
is
R
satisfies
is dense in some line in
C
through
mE
2.
called the period parallelogram of Let
R
satisfies ( 4 ) and (5).
satisfies ( 4 ) and ( 5 ) .
It EIR: 0 < t < 11,
has a positive area.
nw1+mw2= 0
Assume that
and
:{xu~ + Y w Z :Xry
P(R) EP(w1,02)
(where
R
and
n.
is a lattice in C r
Conversely, assume that (6)
is that
Clearly ( 4 ) holds if and only if
( 4 ) but not (5); then
0.
in
is linearly independent over Q r
Proof.
implies
L
E
[ O r l ) I ,
as usual).
R.
Let
P(R)
will be
Note that by (5)
P(R)
88
Norman L. A l l i n g
-
-
P(R) z P ( w 1 , w 2 )
(7)
P(R) i s
then
(8)
[0,111;
E
a compact p a r a l l e l o g r a m a n d h a s v e r t i c e s a t
and
0,w1f~2f
{XUl +yw2: x , y
3
Clearly
w1+02.
(A+P(R))XEL
i s a p a r t i t i o n of
From t h e p i c t u r e i m p l i c i t i n ( 8 ) o n e sees
C.
immediately t h a t
i s a d i s c r e t e s u b g r o u p of
L
proving t h e
C,
theorem.
( 5 ), w i l l be d e n o t e d by
of
T,
R.
t h e q u o t i e n t of
~
(9)
(:) u
=
L.
R
and l e t
@,
B E GL2(Z);
( i i ) Given two b a s e s of
R'.
t h e e n t r i e s of GL2 ( Z )
Clearly C
t h e i d e n t i t y e l e m e n t of t h e s u b g r o u p of L" c L ' c L.
be a b a s i s
E
Since
R"
L.
L,
there exists
B E M
B E GL2(Z).
212
over
Clearly
then t h e r e
R' = B R . C g e n e r a t e d by
such t h a t
R"
Let
R;
thus
(Z)
IR, B
so t h a t R=CR';
R";
L" = L ,
CB=I.
thus
being be
L"
then
proving t h a t
R'
are i n
Similarly there Since
R=CBR.
Similarly
I
and l e t
CR',
R' = B R .
Since
L.
CB = I ,
( i i ) S i n c e t h e e n t r i e s of
C E M ~ , ~ ( Z )so t h a t C
=
is a basis
i s a s u b g r o u p of
GL2 ( Z )
GL2 ( Z ) .
CB = I ,
i s a b a s i s of
i s a b a s i s of
L'
R'
R',
and
g e n e r a t e d by t h e e n t r i e s of
C
R'
exists
R
L,
BRZ
b e t h e s u b g r o u p of
L'
there exists
then
such t h a t
B E GL2(Z)
(i) L e t
Proof.
E
be a l a t t i c e i n
(i) L e t
e x i s t s a unique
B
.r=q(~).
L. Theorem.
of
and w i l l be c a l l e d
Note t h a t
~where
L
Let
6.21
of
,
q(R),
BC=I;
i s u n i q u e l y d e t e r m i n e d by
R
R
hence and
R',
89
The Introduction of Analytic Function Theory proving the theorem.
R
Let
6.22
be a basis of a lattice
quotient (6.20) is
By (6.20:5),
T.
L
in C
Im(.r)# O .
R
whose
will be
called positive or negative according as Im(.r) > 0 or Clearly if fi( = (w1w2)t ) is negative then (ol - w2)' positive basis of B :
Let
(Ei: Eii) q(R');
T'
(2)
T' =
Since
E
L;
thus
GL2(Z)
L
R' EBR;
and let
. B
is non-singular,
It is easy to see that
Im(r') = (Im(-r)detB)/Ibll+b12?I 2
(3)
then
then
(6.20:s) , and since
b l l + b 1 2 ~# O .
is a
always has a positive basis. Let
(b21+ b22T)/ (bll+ b12T)
r LIR
Im(r) < 0.
.
Thus we have the following from Theorem 6.21. Theorem.
B E SL2(Z),
then
positive basis such that
6.23
Assume that
R'
R
is positive.
is a positive basis of
BR
of
L
Given
(1)
L.
(ii) Given a
then there exists a unique
B E SL2(Z)
R' = B R .
Two lattices
L
equivalent if there exists
and
L'
c1 E ~ * ( : Q : -
in c
{Ol)
will be said to be such that
L' =nL.
Clearly this is an equivalence relation between lattices in Let (1)
R
be a positive basis of Let
RT
1
5 (T)
L
and let
let E M ~ , ~ ( C and )
T :q ( R )
L :Z$rZ. T
.
C.
Norman L. Alling
90
L
and
LT
are equivalent, since
Let
6.24
WILT = L.
be a lattice in a: and consider the follow-
L
ing exact sequence in the category of Abelian groupsl where
x-
qyL:
(1)
R
O-M-X~O:
(i.e.l .9 X
a: onto X whose kernal is L).
is a homomorphism of
L
may be topologized by requiring that
be an open mapping
X havethe weakest topology making' L
and that
is easy to see that
continuous.
It
is a compact topological group, which is
X
homeomorphic to a torus. Let lent. (2)
and let
a EC*
Let
X'
f
C/L'
be exact.
Let
5
aL;
then
L'
and
clzl for all
f( z )
z
E
Clearly
C.
onto
X'
Proof.
flL
denotes L
onto
f
restricted to
L.
L'.
making ( 3 ) commutative (1.e. , such that Let
x
E
since (1) is exact. other preimage L.
@.
There exists a homeomorphic isomorphism g
Theorem.
E
is an
R
It is clearly an isomorphism of
h
f
O---,L---*C~X-+O
is, of course, row exact.
X
are equiva-
'-0
analytic autohomeomorphism of the additive group (3)
L
and let
R'
o--;rL'--,B:-x
L'
z1
X.
z
of
f(z1 ) = a z + c c h .
There exists
z EC
x
in
Since
Q:
is of the form
L' =aL,
f(zl)
gR = L'f)
and
x.
z+A, f(z)
.
R(z) =x,
such that
is not uniquely determined by
of
Any for some
differ
The Introduction of Analytic Function Theory L' ,
by an element in
is well defined.
g
hence
R ' (f (2,))
91
.
Thus
g
has
R ' (f ( 2 ) ) 5 g ( x )
=
The reader can easily check that
the requisite properties. As stated this theorem i s not very strong or in-
Remark.
teresting, for groups to
W/Z)
X
and
X'
are isomorphic
as topological
and hence isomorphic to each other.
@ CtR/Z)
The importance of this theorem emerges only when we define analytic structures for is also analytic. and
X
and
then
X';
it turns out
that
g
Analytic equivalence between the surfaces X
is much stronger than topological equivalence, as we
X'
will see.
F i e l d s o f elliptic functions ~-
6.30
Let
(1)
L
be a lattice in
f
F(L) !If:
a meromorphic function on C
f ( z + A ) =f(z),
f E F(L)
Let
C.
for all
z EC
and all
is said to be invariant under
respect
L.
Let
f?:
(w
w2)t
L
function, if
6.31
lytic; then Proof.
L
f
L.
f E F(L) f
w2.
is
may
L-elliptic function or merely an elliptic
is known to be the lattice in question.
is constant.
Let
continuous map of pact, f
L).
and
w1
Liouville's Theorem (1844). Let f
E
such that
or automorphic with
be a basis of
also called doubly periodic with periods also be called an
X
,
R
is bounded on
is bounded on C.
L
and note that
(6.20:7) into C .
P(Q)
-
P(Q).
be ana-
(See (6.12) for historical details.)
be a basis of
-
f E F(L)
Since
Since
-
P(f?)
u XEL ( X + P ( R ) )= C
f
is a is com(6.20:8),
By the classic theorem, usually called
92
Norman L. Alling
"Liouville's Theorem", f
is constant, proving the Theorem.
Since the reader is hopefully very familiar with the classic theorem called "Liouville's Theorem" we have used it to prove the original theorem of Liouville.
6.32
Liouville's theorem has other immediate important
consequences, when applied to elliptic functions. Let
Theorem.
f,g E F(L).
have the same poles on that each pole
of
has
f
corresponding pole of that
f
and
h = f/g.
Then
Theorem, h
6.40
-
on
P(R);
f-g
then
In case (i), let
Proof.
of
and
g
L,
and
the same principle parts as the then
g;
R
for some basis
f
is constant.
(ii) Assume
have the same zeros and same poles - each to
g
the same order
P(R),
(i) Assume that
h
E
F(L)
and
h
f/g
h
G
f
is constant.
- g.
In case (ii), let
is analytic.
By Liouville's
is constant.
Some
a p p l i c a t i o n s of C a u c h y ' s
and
Liouville's
work.
Cauchy's theory of residues can also be applied with considerable effect. of
(1)
L,
f E F(L) t R = (ol w 2 ) Let
and let
R
be a position basis
.
with
A (R) :8 (P(R))
thus it is
Let
,
be positively oriented;
[O,wll u [wl,wl + w21 u [wl + 02,w2] u [w,, 0 1 , oriented
by reading from left to right.
Since
f c F(L)
can have only a
finite number of poles on
F ( R ) , there exists z o
could be chosen to lie in
P(R)) such that
A(n)
f
+zo. Residue
T h e o r e m for e l l i p t i c f u n c t i o n s .
E
C
(which
has no poles on
The Introduction of Analytic Function Theory
(2)
JA(W
93
f (z)dz = 0; +zo
thus the sum of the residues of
f
inside
A(Q) + zo is zero.
Proof.
z +w f (z)dz = A(Q) + zo zO
fzo+w2 f
+w +w f(z)dz
f(z)dz+j20+w1
zO
f (z)dz
Z0+W1+W2
+
Since
! +!
2
f (z)dz. z +w 0 2
is doubly periodic of periods
w1
and
w2,
the right
hand side of (3) equals z +w
z +w f(z)dz+!
(4)
2O
zO
Z
f(z)dz+jZ0 f(z)dz+j f(z)dz, z0+w 1 z +w2 0
which clearly is zero, proving the theorem. Given a function f z
in CC
(5)
let
Resz(f)
(see e.g.,
meromorphic in some neighborhood of
denote the residue of
f
at
z
[69, p. 1111 for the definition); thus ( 2 ) is equiv-
alent to Resz (f) = 0. Note:
( 2 ) is dependent on
f
having no poles on
however the statement of (6) is not. holds for all
(7)
Let
S
zo
A (Q) + zo;
One easily sees that (6)
C.
be a set of coset representatives of
d L
in
then (8)
l z e s Resz(f) = O . Corollary.
Given
f c F(L) - @ ,
then
f
can not have a
@;
94
Norman L. Alling
single simple pole on
S;
thus the degree of poles of
f
on
is at least 2.
S
Note:
(9)
one can prove a version of Theorem 6.32 for
MZ
Let
6.41
S.
denote the set of all functions meromor0
phic in some neighborhood of extension of C .
For
in C.
zo
-
ft-M Z
Clearly it is a field
let
{O},
0
(1)
vz (f) be the unique n
f(z) (z - z ~ ) - is ~
such that
E Z
0
bounded and bounded away from zero in some deleted neighborhood of
v
in
zo
(f) is the order
of
C.
f
at
It is easy to see that
zo.
zO
for all
- {O}.
f , g M~Z
ation on
MZ 0
.
Thus
0
is a discrete rank one valu-
vz 0
(See e.q., [ 7 0 , v01.21 for a discussion of valu-
ation theory. ) Let
Theorem.
(4)
FZES
f E F(L)*;
vz(f) = 0 . Let
Proof.
(5)
ilD(f) :f1/2.rrif;
then
ilD(f),
and
vz(f)
then
the logarithmic derivative of is the residue of
kD(f)
at
f,
is in
z. Hence (4) follows
from (6.40:8). Corollary.
Let
f E F(L)
Let
-
C;
f E F(L) - C.
then
F(L);
For all
a
E
C
The Introduction of Analytic Function Theory
(7)
L
-
S
vz
(f)
N.
is in
(f)
2.
S.
and will be denoted by
It
ord(f).
Clearly
ord(f) =ord(f - a ) =ord(l/f)
for all
a
C.
E
From these remarks we have proved the following.
6.42
Let
Theorem.
f EF(L - C
be of order
n.
For all
aeC L
vz (f - a) = n.
S
vz(f-a) > 0 If
fcC
6.43
R (:(wl f
we will define
Let
Lemma.
w2) t )
f E F(L) - C
f
and
bl,...,b n
0.
be of order
be a positive basis of
has no zeros or poles on
zeros of
ordf to be
A(Q) + z o .
L.
Let
Let
be the poles of
n.
Let such that
zo E C
all...,a n
f
in
be the
P(Q)+zO,
each appearing to its multiplicity: then (1)
- bj) E
r;=,(aj Proof.
L.
Consider the complex number
First let us see that sideration.
A(R) + z o I (3)
IJ
IJ
is related to the problem under con-
Since, by assumption, is well defined.
g ( z ) :zf' (z)/2iiif ( 2 )
f
has no zeros or poles on
The only possible poles of
Norman L. Alling
96
A(Q) + z o
inside
occur at the points
By Cauchy's residue theorem, 1-1
al,
A(R)+zO.
Let
inside
A(n) +
It is easy to see that
zo.
ResC ( g ) =cvc(f);
thus
(5)
P = Ij=l(aj n - bj).
Clearly
f'/2nif
is in
F(L)
.
To help simplify
To help to simplify 1 2 ,
13,
let
5 :z - w 2 ;
( 1 , (2) is
established,proving the Theorem. Apparently by 1736 Euler knew that
([43, pp. 237-2381).
Since
k=m
1/k2 >
1-
From this and (6) we obtain
dx/x 2 21/3m 3 ,
m
obtain m
(')
lk=m
0) > 1/3m3.
Note also that
for each
m
E
N
we also
Norman L. Alling
110
L k ( ~ ) has
(10): : : u
Let
7.32
(1)
zo
Br(zo) : I z
E
E
C:
4m(m-l)
points in it.
r > 0.
C and let Iz
- zoI
0,
is uniformly conver-
B,o.
gent on
Using this Lemma we know, from the classical theory of Weierstrass products (see e.g., [60, vol.1, p . 114 ffl), that
defines an entire function having as zeros exactly the set
L,
each zero being simple. As remarked before, Tannery and Molk [60] take (2) to be the definition of
u
[60, vol.1, p. 1551;
whereas Weierstrass gave (2) [64, vol.V, p. 1201 as an expansion of a known function, a . Since (3
u
(-2) =
il
E
-u
L*
-+
-il E
L*
is a bijection,
(2):
.,
u
is an odd function.
Since
u
is entire and is non-constant, it cannot be doubly
i.
periodic. We want to define Weierstrass’s ?-function as follows.
To make sure that the expression on the right of (4) defines a
111
Weierstrass's Work on Elliptic Functions meromorphic function, let
C-L,
and let
-
of course
a.
E
K
be a non-empty compact subset of
r: supZEK/zI. Let
L'
is finite.
Let
L'
L" 5 L
E
I.L E L: ILI 22r3; then
- L'.
For
z
E
Br(0)
-
and
L",
1
1
1
(5)
1 2- 7 1
=
(2Q-z)z < 3 1 ~ / r - 12r (z-L)2Y21 - (le1/2)21a.12 -
*
Using (7.31:l), together with (5), we see that the right hand side of ( 4 ) converges absolutely uniformly on
K;
thus 'p
having as its set of poles
a meromorphic function on C ,
is L,
each of these poles being of multiplicity 2 and of residue 0. It is clear that (6)
p(-z) = p ( z )
thus
P
zrC;
is an even function.
directly that 'p take
for all
mEL
It is tempting to try to prove
is invariant under
L.
To do this one can
and note that
Rearranging the right hand side of (7) to equal the right side of ( 4 ) would seem to be blocked by the fact that
CLEL*1/E l L The standard way around this diff culty
is divergent (7.31:2).
PI,
is to consider
and to note that
and that the expansion on the right converges absolutely uniformly on
K.
'p'
is
-
of course - meromorphic on
set of poles the set
L,
Clearly
Further, one easily sees that
9' ( z - m )
=
P'
(z),
has as its
each of these poles being of multiplic-
ity 3, and of residue zero.
(9)
@,
'p'
is an odd function.
Norman L. Allins
112
L E L
since
is a bijection, and since the risht hand
L
I l + m E
+
side of ( 8 ) is absolutelv converqent on F(L) - C
'p
E
Let
Proof.
F
(L) -C.
R
( E (wl
j = 1 and 2 .
Let
g(z t
f .( z )
regular at
be a b a s i s of
- ?(z)
for all
f! ( z ) 3
for each thus
L:
Since
j.
and
.p(ztwj)
is a basis
Q
are
'p(z)
Since
-'p(-wj/2).
I
and each
z
and for
z EC,
j i 3 is an even function (6), this number is zero: thus 3
Clearly it
and note that
c = f ( - w./2) ='p(wj/2)
w./2.
L.
3
I
is not in
7
t)
0.)
3 f ( z ) = c . E C, 7 3
is zero: thus w./2
w2)
p ( z + w j ) = p ( z ) for all
suffices to show that
Lr
is in
'p'
(6.30:l).
Theorem.
Of
Thus
K.
.p
f (z) =0 I
7
j, Provins the theorem.
Thus we see that 'p
and
are in
'p'
F(L).
The alqebraic
relation between these functions is of qreat importance to us, and will be established in the next section.
pole in
R :(wl
Let
7.33
at
p(R),
be a basis of
w2)t
0,
0'
The zeros of and
(w,+021/2,
(1)
Let
(w1/2,
Proof.
Let
,
on
E
countinq multiplication. P(Q)
are at
w1/2,
be called the half-periods
(w,+w2)/2,w2/2)
R. be a half-period; then
is an odd function
9' ( - E ) = - P I ( € 1 3
'p'
P(Q),
Each of these zeros is simple.
w2/2.
associated with
Since 'p'
has, bv the same argument, order
has 3 zeros on
Lemma.
has onlv one
and that Dole is of multiplicitv 2; thus
7, is of order 2 (6.41). 'p' 3; hence
L. 'p
p'
(-E)
proving that 9'( E )
it has at most 3 zeros on
P(R).
=-TI = 0.
and
E{L
(E).
'pl
(E)
Since '0'
2~
EL.
= ' p ' ( E - 2 ~ )=
is of order
Since we have found all
three, each must be simple, provinq the lemma.
Weierstrass's Work on Flliptic Functions
113
It is sometimes convenient to define
w - w1/2
(2)
w' !w2/2;
and
Clearly the set of zeros of Let
?\(w1/2)
E
p'
is the set
p( (wl + w 2 ) / 2 )
ell
= (w
w') t ,
and it is a
1 zL.
basis of the lattice
(3)
71 Q
then
Note that the indexing in (3) is not
E
-
e2,
1
L.
zL-
and
p(w2/2)
unfortunately
-
It is in conformity with [ 3 6 ] , but not with all texts.
E
standard. One
justification for this indexing is that with the boundary -
of
e3.
A(R)
positively oriented,the half-periods occur in the order
P(Q)
given in (1) and that the order is compatible with (3), provided
R
is positive.
Of greater importance is that the indexing of
R
(3) is dependent on the choice of basis (4)
The numbers
Indeed,
p'
el,e2, and
is zero at
p(z-w1/2) - e l the value
el
w1/2,
e3
not assume the value
el
01/2.
L.
are distinct.
the zero being simple: thus
has a double zero at doubly at
of
w1/2.
Since
again on
p
P(Q).
That is,
p
assumes
is of order 2 it canA similar statement
holds for the other half periods, estahlishing ( 4 ) . Let (5)
ncN,
sn(L)
with
zPEL*P-"
nZ3,
and let
.
By (7.31:2), the series on the riqht in ( 5 ) is absolutely convergent: thus
s,(L)
complex number.
(or
sn
for short) is a well defined
The right hand side of (5) is known as an
Eisenstein series. (6)
s n = 0,
Indeed, let
CREL*
for all odd n
n L 3 , and let it be odd. - S n = -n thus sn = 0. since -I,* = L*; = s n' (-a)-" = zmEL*m E
N,
n.
114
Norman L. Allinq Let
r:q.l.b.{ILI
REL*);
then
Consider now the Laurent expansion of and let
Q E
L*;
then
lz/.Q
r>O. 'p
Let
about
U-Br(0).
0.
Let
z
E
U
< 1 ; hence
As a conseauence, on rearranqinq we find that
(See e.q., the main rearranqement theorem in [43, PD. 143-1441 Usinq (6) we see that (10) simplifies to
for details.)
the converqence beinq absolutely uniform on compact subsets
u-
(01. Theorem.
(12)
('PI
(z,L))2=4'p3 (Z,L) - 60s4 (L)'p(z,L) - 140s6 (L),
z
\c.
E
Proof.
(13)
'p(z) = z
(14) 'p' (15)
(2)
We have seen (11) that on -2
+3s z
=
4
4 +5s6z +
... .
+ 6s4z + 20s6z3 +
['p' (z)I2 = 4z-6
(16) 4'p3 ( z ) = 4z-6
2
U-
fo)
Thus
... ,
- 24s4z-2 - 80s6 + . . .
+ 36s4z-2 + 6 0 s 6 +
. ..;
hence
for all
W e i e r s t r a s s I s Work on E l l i D t i c F u n c t i o n s
(17)
115
[ p ' ( 2 ) ] - 4p3 ( 2 ) + 6 0 s 4 p ( z ) = - 1 4 0 s 6 + . . . .
Since both s i d e s of
( 1 7 ) a r e e l l i p t i c f u n c t i o n s and s i n c e t h e
f u n c t i o n on t h e r i g h t h a s no p o l e s , w e mav a p p l y L i o u v i l l e ' s Theorem ( 6 . 3 1 ) and c o n c l u d e t h a t t h e r i g h t hand s i d - e o f constant: thus it i s
provinq ( 1 2 ) ,
-140s6,
(17) is
and h e n c e t h e
I t is convenient t o d e f i n e
Theorem.
(18) g 2 ( L )
t o be
and
60s4(L),
q3(L)
t o be
140s6(L);
From t h e Lemma a t t h e b e g i n n i n g of t h i s s e c t i o n , and u s i n g
( 3 1 , w e know t h a t (20)
[P
I
(2)
I
= 4
CP ( 2 ) - e l ) CP( 2 ) - e 2 ) CP( 2 ) - e 3 ) .
On m u l t i p l y i n g o u t t h e r i g h t hand s i d e o f
(201, a n d u s i n g
(20)
and (191, w e see t h a t
(21)
e l + e 2 + e 3 = 0 , e l e 2 + e2 e 3 + e 3 e l = - g 2 / 4 ,
and
ele2e3 = g 3 / 4 . Further, the discriminant, s i d e of (22)
g
o f t h e c u b i c o n t h e r i g h t hand
A,
( 1 9 ) and ( 2 0 ) c a n h e g i v e n a s follows: 3 2
-
2 27g3 = A = 1 6 (e l
(See e . g . ,
By ( 4 1 1
- e 2 )2 ( e 2 - e,)
[ 3 6 , p.
2
(e,
- el) 2 .
1681.)
A # 0.
W e may o b t a i n more i n f o r m a t i o n a b o u t t h e c o e f f i c i e n t s o f t h e Laurent expansion of follows. (23)
let then
'p
about
(ll), by p r o c e e d i n g a s
0,
First t o reduce t h e n o t a t i o n a l burden s l i g h t l y , b ( n ) E ( 2 n + 1)s b(1) Eg2/20
~ and
,
~f o r + n
E
~ N;
b ( 2 ) =g3/28.
Norman L . Alling
116
(24) Let
T(z) =
b ( - l ) 5 1;
and let
h(0) - 0
In--- lb(n)z2n, m
on
then
TJ- {O}.
Differentiating (19), with respect to
z
gives
~ P ’ ( z ) ’ ~ ” (=12p z ) 2 ( z ) ~ ’ ( z-g2P’(z); ) thus
From (24) we obtain m (26) $ o ” ( z ) = ln,-12n(2nl)b(n)z 2n-2 ,
on
IJ- {O}.
The left hand side of (25) is then (27) 6z-4 + 12b(l) + r:=22n(2n-l)h(n)z
2n-2
.
The right hand side of (25) is
(29) thus, For
(29)gives
n = 2,
Note that
n(2n-l)h(n)
(30) b(n)
=
31y=-1b(j)b(n-1-j),
6b(2) = 6h(2),
1123,
3(l~~:b(j)b(n-l-j))/(n-2) (2n+3).
b(n) ,
n 2 3.
for each
expression concerning the (31)
~
for
m24.
Let
L
and
L’
g . (1,) = g . ( L ’ ) ,
3 I Proposition.
b(2),
s
(2k-1)(2m-2k-1)~
j= 2
Assume that
and
Converting (30) back into an
be lattices in for
b(1)
~ gives ~ ‘
(2m + 1) (m-3)(2m-1)s~~ = 31:1;
implies
n22.
and thus is uninteresting.
Clearly (30) is an algorithm which, given determines
for each
+ b(O)h(n-l) + b(n-l)h(O) + b(n)b(-1) = 2b(n) ;
b(-l)b(n)
thus for each
=
and
g .
3
Clearly
@.
~
L=L’
3.
(L) = g .( L ’ ) 7
for
j=2
,
~
Weierstrass's Work on Elliptic Functions and 3; then
117
L=L'. Using (30) we see that the Laurent expansion of
Proof.
'p(z,L) and of
at
T(z,L')
0
are identical. Hence
rp(z,L) =
'p(z,L'). As a consequence their period lattices are equal, proving the proposition.
7.34
Weierstrass introduced the function, that was later
known as the Weierstrass
zeta function
O . Let
and let
Since
converges (7.31:l) , we see that the series
lREL* I R
on the right hand side of (2) converges absolutely uniformly on K;
thus it can be used to define
meromorphic function on Q: at the points of
L.
P(z)
=-
d 0;
Assume that and
g
then
w o ( f ) < 2m.
129
By construction
have exactly the same zeros to the same order on
P(Q) - I O l .
Let
bl,... ,br
be A-points of
so that
P(Q)
{bl,bi,...,br,b~~ is the set of all zeros of
f
in
P(Q)
are A-points, each indexed according to multiplicity. again that (11) holds.) P(Q)
Let
in
f
that are half-periods, each occurring to half the mul-
constant.
Since, by construction, q
it must have a pole at on
P(Q) - { O l Hence
f
in
has no poles on
thus the zeros of
0;
f,
and
listed above are all the zeros of
2(r+s) =q.
By construction q
is not P(Q)
- {O}
hence of q
on
By (20)
. . . + cqpq
(24) q = co + clp +
(Note
P(Q).
Since, by assumption, q > 0, q
again that (15) holds.)
P(R).
that
(Note
be the zeros of
61,... , 6 s
tiplicity of the corresponding zero of
g,
f
E
c "p1 , with
cq # 0 .
and
have the same zeros and the same poles to the same orders on P(Q);
thus by Liouville's Theorem c
(26) f =
g = c t; 9
hence
nj,lr ('P - P (b.1 1 fl Sj = l ( P - 'P (6j ) 1 k n njz1 ('P - ?(aj) ) I ' I ~ , ~( P - ' P ( c j ) )
(Note that (23) may be regarded as a special case of (26) in which
r = O = s.)
By construction, the rational function given
in (26) is reduced.
In terms of
k,nrrr and
s,
w0(f) =
2(k+n) -2(r+s). Bibliographic note.
The Theorem above is asserted by
Weierstrass [64, vol.V, pp. 141-1521.
A
formula quite similar
to (26) can be found in Whittaker and Watson [69, p . 4491.
A
Norman L. Alling
130
more detailed treatment may be found in [36, pp. 171-1721.
Let
7.43
f E F(L)
have only simple poles, let
.
bll.. ,bn be the poles of residue of
f
at
b
f
P ( Q ),
151 'n.
for
j'
in
and let
r
j
be the
We have seen ( 6 . 4 0 : E )
that
(2)
Let
then
q
g(z
+ R)
=
q(z)
: I n1=1r.c(z-bj); 3
is a meromorphic function on @ . Given R E L, n equals lj=lrjc(z + R - b . ) , which by ( 7 . 3 4 : 7 ) 3
l;,l.rjc(z
- bj) + I;,lrjcR
g E F(L).
Since
C,
f-q
which, using (1), equals is in
it is a constant c.
(3)
F(L)
g(z) ;
thus
and is analytic on all of
Hence
f(z) =c+lj,lrj5(z-bj). n
This method of representing
f
can be used to prove an existence
theorem, namely the following Let
Theorem.
tinct points in
(5)
Let
nENI n22,
P (fl)
.
Let
r
j
EQ:
and let bl,...,b n be dis*, with 15 j 'n, such that
f(z) ~ ~ + ~ ~ = ~ r ~ < ( zfor - ball , ) Z, E C , 3
where
c
E @ .
its poles in
Then
f E F(L) ,
f
P(fl)
being at
bl,...,bnI
has only simple poles on C I
.
rll.. ,rn'
An a d d i t i o n t h e o r e m
7.50
Let
u
E
c
-
1 -L 2
for
and consider
'p
with residues
Weierstrass's Work on Elliptic Functions
f (z) 5 p ( z )
(1)
-
P(u),
for all
131
zE C .
f
has a double pole at each point of L, and has a root at df each point of u + L. Since u P T1L , =(u) = 7' (u)# 0, showing that each point in
u+L
is a simple root of
f.
Since
an even function, f (-u)= p(-u) - p ( u ) = I)(u) - I)(u) = 0; f(z) and u(z-u)u(z+u)/u 2 (z) are both elements of
p
is
thus F(L)
that have the same zeros and same poles to the same orders. From Liouville's Theorem we conclude that there exists
cEC*
such that (2)
f(z) =ca(z-u)u(z+u)/a
(3)
Let
2
(2).
g(z) -cz2a(z-u)u(z+u)/u 2 ( z ) ,
for all z E C . g is an entire function and g ( 0 ) = cu ( - u ) u (u)= 2 -CIS (u) (since u is an odd function). From (1) we see that the principal part of (4)
c = -l/a
(5)
Hence
2
f(z)
at
0
is
z - ~ ; thus
.
(u)
p(z) - p(u) = - u ( 2 - u ) 0 2(z+u) 02(z)o (u)
Note that ( 5 ) was established for all
zEC
and all
u
E
C
1 - 2L.
By the identity theorem for meromorphic functions, ( 5 ) holds for all in (6)
z
z
and
and
u E C.
Let
h(z,u)
u.
ah 6Z(h(z,u)) --(z,u)/h(z,u) az ah 6U (h(z,u)) :-(z,u)/h(z,u). au Let
Applying
be analytic and non-zero
E Z to both sides of
(Note that ( 7 . 3 5 : 3 ) holds for
I
and let
( 5 ) gives
-
Dz.)
Applying
-
DU
to both sides
132
Norman L. Alling
of (5) gives
Adding (7) and ( 8 1 together, and dividing by 2 , gives
Differentiating (9) with respect to
z
gives
The left hand side of (10) is
thus
Interchanging
z
and
u
in (12) gives
Adding ( 1 2 ) and (13) together and dividing by 2 gives
(7')2
Since
=4?
3 -g2'p - g 3 ,
(7.33:19),
r ) " p p " = (1r)2 - g 2 ) ? ' ,
and hence
Using (15) to simplify (14), we obtain Weierstrass's Addition Theorem
Let
u
be a point in s: - L
and let
z
approach it;
then (16)
Weierstrass's Work on Elliptic Functions
133
gives
(16) may be found in Weierstrass's
Bibliographic note.
lectures [64, vol.V, p. 381. p. 2181.
(17) appears in [64, VOl.Vr
The treatment given in this section was found by the
author in [54, pp. 384-3861, and modified slightly.
A
7.60
(1)
relation between Weierstrass's
2 f (v) E exp (-nlwlv /2) u (w,v)
Let
simple zero at 0(
+w.
z
for all (3)
function
and
,
v ~ c . Clearly f is an entire function, which has a
for all
(2)
0
=
7
z
0.
Recall (7.35:13) that
-exp ( q . ( z 3
and
E @ ,
j =1
+ w 7./2)1 0 ( 2 ), and
2.
f (v + 1) = -f (v), for all
.
E
C.
w 2/ w 1 (5.20:l) , and that k = exp(-in.r - 2niv) Using Legendre's equation (7.34:11) one can show
Recall that (5.30:1)
v
Then
T 5
that (4)
f(v+T) =-kf(v),
By Theorem 6.54, c
for all
f (v) = cel (v),
VE@.
for some
cei(0) =f'(O) =w,a'(O).
note that
c E C.
To evaluate
We can compute
5'(0)
directly, by differentiating the product (7.32:2) used to define it, and evaluating 0' ( 0 )
= 1;
thus
u'(v)
at
c = w 1/ e l1 ( 0 ) .
( u l / O i ( 0 ) )exp(nlwlvL/2)e,(v).
(5.20:3), let (5)
0 (u) =
u:olv.
0.
On doing so we find that
Hence
0 (w
1v) =
Using the standard notation,
Then
.
2 (wl/O; ( 0 ) 1 exp (qlwlv /2) el (v)
This Page Intentionally Left Blank
CHAPTER 8 RIEMANN SURFACES
Introduction
8.10
Bernhard Riemann (1826-1866) Galois,
... m a t u r e d
,
l i k e Gauss, A b e l , E i s e n s t e i n ,
mathematically a t an e a r l y age.
H e worked a t
t h e v e r y h i g h e s t l e v e l from a b o u t 1 8 5 0 , u n t i l s h o r t l y b e f o r e h i s d e a t h , a t t h e a g e o f 4 0 i n 1866.
H i s main work o n w h a t came t o
der
be known a s Riemann s u r f a c e s was h i s T h e o r i e
Abel'schen
F u n c t i o n e n , which a p p e a r e d i n C r e l l e ' s J o u r n a l i n 1857 [ 5 3 , pp. 88-1421.
While t h i s work w a s v e r y i n f l u e n t i a l , i t w a s d i f f i c u l t
t o r e a d and absorb.
A g r e a t d e a l of e f f o r t h a s been devoted t o
e x p l o i t i n g t h e p o t e n t i a l t h a t l a y i n t h e s e i d e a s o f Riemann. During t h e 1880-1881 academic y e a r a t G o t t i n g e n F e l i x K l e i n l e c t u r e d o n Riemann's Theory o f a l g e b r a i c f u n c t i o n s a n d t h e i r F o r t u n a t e l y t h e s e l e c t u r e s were p u b l i s h e d .
integrals.
There
i s e v e n a n e x c e l l e n t E n g l i s h t r a n s l a t i o n of K l e i n ' s l e c t u r e s [39].
D u r i n g 1 9 1 1 - 1 9 1 2 a t G g t t i n g e n Herman Weyl a l s o deThese a l s o a p p e a r -
l i v e r e d a series o f l e c t u r e s on t h e s u b j e c t . e d i n p r i n t i n German a n d i n E n g l i s h
[67].
For a d i s c u s s i o n
o f t h e v a s t c o n t r i b u t i o n s o f Riemann see e . g . ,
Klein's lectures
[401. 8.11
Let
(7.32:4) ;
A n Example
L
be a l a t t i c e i n
C
t h e n a s w e have s e e n , 135
(6.20).
F
Let
(-F(L))
9(z)
:v ( z , L )
(6.30:l)
equals
136
Norman L . A l l i n g
,
(7.42)
@((n,ll)
3
= 4p
(1)
and -g21\-g3,
(7.33:12 y :p ' ;
and l e t
x 5 'p
(2)
Let
(3)
3 y 2 = 4x - g 2 x - g 3 : f ( x ) ,
and 1 9 ) .
then F = @(x,y).
and
n
Further, y
L
are d i s t i n c t complex numbers t h e r e are complex numbers
w?(z) = f ( z ) , 3 Some
k,
,
= 4 ( x - e l ) ( x - e 2 ) ( x - e,)
(7.33:4).
w,(z)
w (z) = 0 = w 2
t h e Riemann s u r f a c e of
y
and
and
z
e3
C
E
such t h a t
W,(Z)
z = ekl
If
for
The c l a s s i c way t o c o n s t r u c t
(2).
1
= f (x)
elIe2,
For every
w2(z) = -w,(z).
and s u c h t h a t
then
where
,
which might be c a l l e d t h e
" c u t and p a s t e method", i s t o l e t
b e t h e Riemann s p h e r e
@'
minus n o n - i n t e r s e c t i n g c l o s e d J o r d a n arcs j o i n i n g el t o
C
e2
e3,
to
e3
and
to
@'
Over
compact s u r f a c e .
then
m;
e 2'
i s a c o n n e c t e d non-
C'
it i s p o s s i b l e t o choose s i n g l e
v a l u e d a n a l y t i c b r a n c h e s of
w1 ( z )
and
and
j = 1
w,(z) ;
t h e n one c a n
form (4)
W' :{ ( z , w . ( z ) ) : z
Let
vl
3
be t h e f i r s t p r o j e c t i o n of
W',
maps
(5)
Let
then
W'
i n a two-to-one
w0 c
W
0
,
and
z
E
0 C
. -
Let
IT
1
injective.
0
T ~ ( W) Pro
and l e t
W
z
@;
c
C
2
T
1.
vl
then
Ct.
Let mW
W
be t h e one p o i n t
denote t h e element i n
by mapping
mW
to
m
t h e r e a r e e x a c t l y two p o i n t s o f under
.
= f ( a ) )I ;
= C.
extend t o
{elle21e31m)
p r o j e c t down t o
B~
and 2) onto
C2
fashion, onto
c2:
:{ ( a , ~ )E
compactification of W-W
@'
E
Over
E
C.
If
W
that
l e l , e 2 , e 3 , ~ ) , -ir1\w
I t i s n o t d i f f i c u l t t o see t h a t
is
Riemann S u r f a c e s (6)
i s a s u r f a c e and t h a t
W
137
i s a n open c o n t i n u o u s map o f
n, I
onto
PJ
C.
C l e a r l y t h e l o c a l behavior of
el,e2,e3,
and
a t t h e p o i n t s above 1 i s e q u i v a l e n t t o t h e behavior of z
m
7~
The E u l e r c h a r a c t e r i s t i c o f
0.
follows.
Consider a t e t r a h e d r o n
el,e2,e3,
and
Clearly
a.
h a s 6 e d g e s and 4 f a c e s : t h u s
T
genus of
is
g(T),
TI
0.
x(T) = 4
(See e . g . ,
ence t o t h e topology of s u r f a c e s . )
w e w i l l b u i l d a geometric f i g u r e t i n u o u s map morphic.
let
p
of
onto
S
T
Over e a c h v e r t e x of
-6+4
Further,
C.
Thus t h e
= 2.
[51] f o r a g e n e r a l r e f e r -
I n o r d e r t o compute
x(W)
over
T
and a n o p e n con-
such t h a t
S
and
S
a r e homeo-
W
c o n s t r u c t a v e r t e x of
T
and
S
map t h e s e new v e r t i c e s t o t h e c o r r e s p o n d i n g o l d o n e s .
p
Over e a c h e d g e o f
T
c o n s t r u c t two new e d g e s of
S
and l e t
map t h e s e new e d g e s down t o t h e c o r r e s p o n d i n g o l d e d g e s o f Over e a c h f a c e o f
T
c o n s t r u c t two new f a c e s o f
ma P t h e s e down t o t h e c o r r e s p o n d i n g o l d f a c e s o f wa Y t h a t a t t h e v e r t i c e s o f
x(S) W
X : @/L
Let
?\
Since
(8)
= 4 - 1 2 + 8 = 0;
a r e homeomorphic,
(7)
X
S,
p
thus g
g ( s ) = 1.
(W) = 1:
S
i.e.
,
T.
p
i n such a
z
z2
+
at
W L > Z . )
Finally, since
W
p
and l e t
TI
is locally l i k e
(To see t h a t t h i s c a n b e d o n e c o n s i d e r
0.
at
whose v e r t i c e s a r e l a b e l e d
T
i s homeomorphic t o
T
z2
may b e computed a s
x(W),
W,
+
and
S
is a torus.
(6.24).
i s i n v a r i a n t under
L
( 7 . 3 2 ) i t i n d u c e s a map
p
of
X.
onto
7T
7\ X->C
is topologically equivalent t o
W-
>C
I n s e c t i o n 8.20 w e w i l l p u t a n " a n a l y t i c s t r u c t u r e " on and o n
C
so t h a t w e may d o a n a l y t i c f u n c t i o n t h e o r y on t h e s e
X
138
Norman L. A l l i n g
surfaces. 8.12.
Let
f (X,Y)
(1) L e t
and
X
be a n i r r e d u c i b l e p o l y n o m i a l i n
n > 0
degree
(in
C l e a r l y we can f i n d (2)
a(X)f(X,Y)
g(X,Y)
( W e can even choose Let
then
I
such t h a t
i s of degree
in
n
Y,
and i s i n
where
g(X,Y) = i jn= o p j ( X ) Y j ,
a(X)
i s a non-zero,
pj(X)
E
@[XI,
x :h ( X )
Let
t o b e monic and o f minimal d e u r e e . )
proper i d e a l i n
i s maximal.
I
@ ( X ) [Yl
morphism of
onto
and
F = @ ( x , y ),
then
where
f (X)@(X)[Yl;
1
is irreducible,
(4)
of
@ ( X ) [Y]
pn(X) # 0 .
and
(3)
@[XI
E
QI'.
n f(X,Y) = l j = o r j ( X ) Y J r
thus
Y);
a(X)
Hence
@[XI [Y].
be two i n d e t e r m i n a n t s over
Y
and
C(X) [Yl/I
h
f (X,Y)
be t h e c a n o n i c a l homo-
: F.
y :h ( Y ) ; f (x,y) = 0 = g(x,y)
how t h e Riemann s u r f a c e
of t h e equation
W
g(x,y) = 0)
equivalently of
Let
Since
@(XI [Y].
.
W e w i l l now s u g g e s t f(x,y) = 0
(or
c a n b e c o n s t r u c t e d by t h e " c u t and
p a s t e method". T h e "cut and p a s t e -
8.13
surface
(1) L e t
E
(2)
F = @ ( x , y ) by r e a s o n i n g a s f o l l o w s .
of
c0
{A€@:
pn(X) # 0
Since A
W
m e t h o d " c o n s t r u c t s t h e Riemann
pn(A)#ol.
(8.12:2),
cc -
C0
is a f i n i t e set.
For a l l
Co
i s a polynomial of degree
l;=opj ( A ) t J
As such it has
n
roots
tl(A),
occurring t o its multiplicity.
. .. , t n ( A )
E
n C,
in
@[tl.
each r o o t
Riemann Surfaces
Let
(3)
Il
D(A) Z
jX
+
-
z1
b e d e f i n e d so t h a t
X
0
l/m
where
Example 1 .
and l e t
U2 : Z
Let
@.
z2
h
U1 : @
Let
0
Zk
-1
I z7.( U7. ) n
zk(Uk)
i s a n a l y t i c , f o r each
(Here t h e empty mapping i s t a k e n t o b e a n a l y t i c . ) ( 4 ) a r e c a l l e d t h e t r a n s i t i o n f u n c t i o n s of Lemma.
analytic.
j ,k
E
J.
The maps of
U.
The A t l a s e s g i v e n i n Example 0 a n d Example 1 a r e
Norman L . A l l i n g
142
U : ( Vk , w k ) kcK
and
U
Let
b e A t l a s e s on
They w i l l
X.
be c a l l e d a n a l y t i c a l l y e q u i v a l e n t i f
j'
(5)
0
-1 Wk Iz. ( u . ) n 3 7
and a l l
k
is analytic, for a l l
J
E
A n a l y t i c e q u i v a l e n c e between a n a l y t i c A t l a s e s
i s an equivalence r e l a t i o n .
X
X
An e q u i v a l e n c e c l a s s l y t i c Atlases of
of a n a l y t i c a l l y e q u i v a l e n t ana-
i s c a l l e d a n a n a l y t i c s t r u c t u r e on
X
pair
(X,X)
c o n s i s t i n g of a s u r f a c e
ture
X
X
on
i s c a l l e d a Riemann s u r f a c e .
n o t e d by
Clearly
C.
Although i t i s a n
i n s t e a d of
(X,X).
i s i n an a n a l y t i c s t r u c t u r e
S
i s c a l l e d t h e Riemann s p h e r e , and i s u s u a l l y de-
(1,s)
1.
U
(Cont.)
Example 0 .
X
The
X.
and a n a n a l y t i c s t r u c -
X
abuse of n o t a t i o n w e f r e q u e n t l y use
on
j
K.
E
Theorem.
of
wk(vk)
i s a compact s i m p l y c o n n e c t e d Riernann
C
surface.
X.
ture
U
(Cont.)
Example 1.
i s contained i n an a n a l y t i c s t r u c -
is an a n a l y t i c t o r u s .
(X,X)
S i n c e a n a l y t i c maps p r e s e r v e o r i e n t a t i o n and s i n c e
C
is
o r i e n t e d w e have t h e following: Riemann s u r f a c e s a r e o r i e n t e d .
Proposition.
Let
8.21.
let
f:X
U E ( U
x
E
+
z )
j f j jcJ
E
X
there exists
X
f (x)c V k f
(1) wk
b e Riemann s u r f a c e s and
(Y,Y)
w i l l be c a l l e d a n a l y t i c i f t h e r e e x i s t s
f
Y.
and
(X,X)
U z (Vk f W k ) k E KE Y
and
j
E
and
J
k
K
E
such t h a t f o r a l l
such t h a t
x
E
U
jf
and o
f
Lemma.
0
z
-1 j
is analytic a t
Assume t h a t
f
z.(x) 3
E
@.
i s a n a l y t i c ; t h e n f o r any c h o i c e
Riemann S u r f a c e s
of
U
E
X
and
V
Y
E
143
(1)holds f o r a l l
j
and a l l
J
E
k
K.
E
This holds s i n c e t h e composition of t w o a n a l y t i c
Proof.
f u n c t i o n s i s a n a l y t i c and t h e i n v e r s e o f a n i n j e c t i v e a n a l y t i c function is analytic. d e n o t e t h e s e t o f a l l a n a l y t i c maps o f
F(X)
Let
o t h e r t h a n t h e map t h a t t a k e s e v e r y
C,
Clearly
F(X)
contains
@.
Wk( t )
(2)
E
m
E
1.
I t w i l l b e c a l l e d t h e f i e l d of a l l meromorphic
X.
x
E
U
0
f
0
j (1) z . J
n U
and
j( 2 )
f(x)
E
Vk(l)
its derivation
for
g;,
t = 1
-1 wk(2)
91 = w k ( l )
n V
Since
k(2)*
-1 = ( t ) - 9t
i s a n a n a l y t i c f u n c t i o n on a n open s u b s e t o f
(3)
to
X
into
i s a f i e l d , under point-wise o p e r a t i o n s , t h a t
f u n c t i o n s on Let
x
X
g2
and
Clearly
2. 0
'j(2)
2
w e can take
@,
-1 j (1)'
Is t h e r e a n y r e a s o n why (4)
gi
o
zj(l)
should equal Let
Example.
U : (U
Let
z,(X)
wl(A)
and
I
A
E
= X I
21 # 9;
0
where
X
X
E
o
z ~ ( ~ ) ?
Let
f(X) :
U1
E @ :U 2 ,
V
Let
@.
for all
for all
g 2 ( A ) = X/2, 0
X :@ E Y .
j l z j )j r 1 1 , 2 1
for all
= 21,
g;
E
@;
@.
(Vltwl),
gl(X)
thus
=
A,
A
for all
z,(A)
=
E
X,
and
=
v1
where for all X
gi = 1 # 1 / 2 = g ; .
E
@.
@,
@.
Hence
22-
Thus t h i s s t a b ( 4 ) a t d e f i n i n g a d e r i v a t i v e o f f a i l s to define an invariant object.
f
on
x
It t u r n s o u t t h a t deriva-
t i v e s a r e n o t d e f i n a b l e o n Riemann s u r f a c e s ; however d i f f e r e n -
t i a l s are, as w e w i l l see i n t h e n e x t s e c t i o n . 8.22.
Let
(X,x)
b e a Riemann s u r f a c e a n d l e t
144
u
Norman L. Alling
(U.,Z.)
f
I
I jEJ
(1) Tjk where
x.
E
D(zj
o
For all
zkl)
z k,
o
J,
E
on
U . n Uk,
1
denotes differentiation in
D
let
j,k
Let
@.
(&j)j E
J
:6u
be
such that
Aj
(2)
is a meromorphic function on
U
(i.e.,
j
6.:U
i
C
-f
j
is analytic and not the infinite constant map), and 6k = 6 . T 1 jk'
(3)
6u
on
is then a meromorphic differential relative ( C o n t . f r o m 98.20).
Example 0. (E @ - ( 0 ) ) .
Let
T21(X)
:1
cS1
on
=
- 1/X 2
2
Note that
L S ~ ( X ) :-A
-2
on
U.
n U2
U1
T12(X) = - X 2 ,
and
and let
U1
(Sj) j E 1 1 , 2 1
sees that
= @*
X
for all
E
@.
One easily
U2.
is a meromorphic differential relative
u.
to
Example 1.
j
U . n Uk. 3
E
-1 z k (1) = X
J.
D(zj
(Cont. from 5 8 . 2 0 ) .
Since z j o zk-1) = 1, and so
o
T
jk
+ =
wjk,
1,
Let 6 1, for all j where 2 w j k E L,
for all
j,k
is a meromorphic differential with respect to In general
6u
E
J.
Thus
6u
U.
is called analytic if each
sj: u j
@.
-P
Note that the differential constructed in Example 1 above is analytic, whereas the differential constructed in Example 0 is not.
Let
yu
be a meromorphic differential relative to
.
Let
f
(4)
(6 + Y . ) I i jEJ
E
F(X)
a
and
with respect to
(f&j)jEJ are meromorphic differentials U.
These differentials will be denoted by spectively.
U.
Note also that
AU 'yu
and
f6U re-
6 j/yj IUj n Uk = 6 /y I U , n Uk; k k l
Riemann Surfaces
145
thus is a meromorphic function
(5) ujEJ6j/yj
is non-zero.
yu
g
X I provided
on
g
will be written as
(6) dfj
D(f
z-1 .)
o
Finally let
GU/yu.
for each
z . 3,
o
3
j
J.
E
is a meromorphic differ(dfj)jE which will be denoted by dfU.
It is easy to verify that ential relative to
U,
Let
8.23.
V
X
E
(1)
and
6u
and
V
and
U
respectively. are equivalent if
yy
X.
D(X)
Let
X.
entials on
is a meromorphic
yy
X
is an equivalence class
6 u with respect to
of meromorphic differentials E
6u u
U u V.
A meromorphic differential on
U
and
be meromorphic differentials rela-
yu
differential relative to
6
U
(X,X) be a Riemann surface, let
6u
and let
tive to
b U = dzU.
(Cont.)
Example 0 .
U,
with
denote the set of all meromorphic differ-
It is easy to see that (8.22:4) induces the
structure of a vector space on
D(X)
over
F(X).
From (8.22:5)
we see that (2) D(X)
is of dimension at mo.st one
From (8.22:6) one (3)
d:f
E
F(X)
kernel is Theorem.
sion of
D(X)
let
bU
E
6
6.
df
i
D(X)
is a C-linear derivation, whose
C.
If there exists
0 # df E
F(X).
easily sees that
over
Proof.
Let
-+
over
D(X)
F(X) E
is
U(X).
with
6
f
E
F(X) - C ,
then the dimen-
1.
Using (2) the theorem is proved.
# 0.
Let
U
( Uj IZj’j E J E X
and
Norman L . A l l i n g
146
Lemma. ____
No
Proof.
Since
6
i s i d e n t i c a l l y zero.
j
6 # 0
i s not i d e n t i c a l l y zero.
there exists
x
Since
k'
i s , by assumption, c o n n e c t e d ,
and s i n c e (8.22:3) h o l d s , and s i n c e e a c h zero, each
6
k
Let
Z.
T
(8.22:l) i s n e v e r
jk
i s n o t i d e n t i c a l l y z e r o , p r o v i n g t h e Lemma.
j
The o r d e r o f in
6k,
such t h a t
J
E
6
a t some x E U j l ~ ~ ( j s u c h t h a t x E Uk. S i n c e T
J
E
6( 6 .~4 1 :) 1 ) ,
is
is analytic
jk
and w i t h o u t z e r o s o r p o l e s ,
(4)
vx(6.) =
Let
~ ~ ( 6d e) n o t e t h e i n t e g e r i n ( 4 ) .
Vx(6) n
b
X
X
divX
E
if
be
X.
Clearly
2
b(x),
for a l l
x
E
{Xx:
x
supp(a),
XI
E
i s a f r e e b a s i s of
t h e s u p p o r t of
deg(a)
Let
Xx(y) = 6 x y l
is
a,
ix
for a l l
divX Given
X;
then
x,y
a
divX.
Given
X:
a ( x ) # 01.
E
deg.
divX
Divisors i n
Clearly
divX/div X
for a l l
x
0
E
X.
E
divX
Ixcsupp(a a )( x ) I
i s a homomorphism of
k e r n e l of
X.
E
is finite.
where t h e sum o v e r t h e empty s e t i s d e f i n e d t o b e z e r o . deg
XI.
i . e . , i t i s a Z-module.
a(x)
such t h a t
By d e f i n i t i o n , s u p p ( a ) (4)
and
X
E
let
Clearly (3)
XI
g,
i s a p a r t i a l l y o r d e r e d ( e v e n a l a t t i c e - o r d e r e d ) group.
divX
(2)
let
divX,
r.
and
w i l l be c a l l e d d i v i s o r 0”
divX
i s a t o r s i o n f r e e Abelian group: E
Pl
i s z e r o e x c e p t on a f i n i t e s u b s e t of
a
Z:
and
P1,
d i v i s o r class g r o u p
t h e group of d i v i s o r s on
(1) { a : X
a,b
A
at
b e a compact Riemann s u r f a c e o f genus
X
Let
r
i s tangent t o
i s a d o u b l e p o i n t of i n t e r s e c t i o n of 8.60.
163
X
Since
Let
2.
divOX b e t h e
divOX w i l l be c a l l e d homogeneous.
Given
Z.
2
onto
Clearly
f
E
i s compact
F(X)* f
let
( f ) (x)
f
vx(f),
can have o n l y a f i n i t e
number of z e r o s o r p o l e s ; t h u s
(5)
f
E
F(X)*
+
(f)
divX
E
i s a homomorphism of t h e m u l t i p l i c a t i v e group a d d i t i v e group (6)
into the
divX.
divX/(F(X)*) 5 C ( X ) Lemma. _-
F(X)*
(F(X)*)
c
i s t h e d i v i s o r c l a s s group of divoX.
X.
164
Normal L . A l l i n g
F7e w i l l p r o v e t h i s lemma i n c a s e
or
g = 0
The
1.
p r o o f i n t h e g e n e r a l c a s e may b e found i n t h e l i t e r a t u r e on t h e subject (7)
.
d i v o X / ( F ( X )* )
i s t h e homogeneous d i v i s o r c l a s s
E Co(X)
We w i l l compute
i n case
Co(X)
g = 0
and
1
i n the
n e x t two s e c t i o n s . See e . g . , X
g r o u p on
[29]
f o r c o n n e c t i o n s between t h e d i v i s o r c l a s s
and c e r t a i n s h e a f cohomoloay g r o u p s on Assume
8.61.
then w e have seen t h a t
g = 0;
a r e analytically equivalent (8.46).
Co(C)
Theorern.
=
Let
a
that
a # 0.
Let
S :s u p p ( a ) .
then
g
and
S'
point i n
E
and
F(Z)*
supp(b)
the points i n
d i v 1. 0
If
Let
b
then
a = ( 1 ) . Assume
n :a(m)
b :a
does n o t c o n t a i n
a.
+
and l e t
is i n
(g)
b
g + zn;
div C 0
xl,...,x
Let
be t h e
k yl, . . . , y k
i s p o s i t i v e , and l e t
o n which
S'
C
( F ( L ) * ) c divoC.
a = 0
( 9 ) = nXm.
on which
S'
and
COI.
Proof.
E
X
W e have seen ( 8 . 3 0 ) t h a t
I n ( 8 . 3 0 ) w e a l s o see t h a t
F(C) = C ( Z ) .
X.
be
i s n e g a t i v e , each o c c u r r i n g
according t o i t s m u l t i p l i c i t y . (z-xl)
(1) L e t then
f (z) :
( f ) = b, 8.62.
Let
(z-y,)
.. . . ..
and h e n c e L
(2-x,)
(z-y,)
E
F(C)*;
(f/g) = a ,
be a l a t t i c e i n
p r o v i n g t h e Theorem.
C
and l e t
complete w i t h t h e induced group s t r u c t u r e . o f genus 1.
By Theorem 6 . 4 1
X
is
X
2
-
of c o u r s e
C/L,
-
165
Riemann S u r f a c e s
(1)
(F(X)*)
(2)
Given
(3)
Define
divOX
c
a
divoX,
E
sum(a)
a = Cxcsupp ( a a ( x ) X x -
t o be
CxEsupp(a
C l e a r l y sum i s a homomorphism o f
a ( x )x .
divOX o n t o
X.
By Theorem
6.43 (4)
(F(X)*)
k e r sum.
c
By Theorem 7 . 4 1 , (5)
given
a
( f ) = a;
E
k e r sum,
i.e.,
there exists
f
E
F(X)*,
such t h a t
( F ( X ) * ) = k e r sum.
Thus w e h a v e p r o v e d t h e f o l l o w i n g . Theorem.
onto
X.
sum i n d u c e s a n isomorphism, Sum, of
c,(X)
This Page Intentionally Left Blank
CHAPTER 9
THE ELLIPTIC MODULAR FUNCTION
Introduction
9.10
We have seen that parameters occur in elliptic integrals. For example, Legendre's modulus Chapter 1. functions
Two parameters, c 54.1.
k
occurs in some integrals in
and
e l occur in Abel's elliptic
Legendre's modulus again occurs in Jacobi's
elliptic functions 94.2.
Clearly
(5.20:l) is a parameter
T
that plays a role in defining theta functions Further, a lattice and
w2,
and
L
- in
L
in C
in Chapter 5.
is determined by parameters
turn - determines
1\
and
O1
7'.
One way to describe the problem at hand is the following: find a complex number associaked with lent to
p(X),
X -C/L,
X'(x/L')
the modulus of
such that
if and only if
X
XI
which can be
is analytically equiva-
p(X) =p(X').
This problem
is addressed and solved by using the elliptic modular function J,
and is the subject of this chapter.
eventually, that
J
We will also see,
is intimately connected with the parameters
mentioned in the paragraph above. Klein traces the origins of the elliptic modular function back to work of Gauss.
(See [40, p. 43 ff.].)
Klein himself
did a great deal of work on the elliptic modular function.
(See
the third volume of his collected works and the huge two-volume work, written with Fricke, on the elliptic modular function 1411.) 167
168
Norman L. Alling Finally note that "modular" was used in one sense prior
to about 1850 and in another after about that date. Weierstrass remarked [64, vol.1, p. 501
As
in a note in his col-
lected works written near the end of his lifeton his first papers, the term "modular function" c. 1840 referred to what we now call "elliptic functions". Let
9.11
be a lattice in c ,
L
let
n e N,
with
n> 3,
and recall (7.33:5) that the Eisenstein series (1)
L&-n
Zt
converges absolutely and is defined to be
sn (L) Let
a
E
c*
and let
lattices in c (2)
L' :aL; then
(8.45).
are equivalent
Clearly
g2(L) = 6 0 s4(L)
and that
g3(L) =
thus -4
g2(L') = a
g2(l)
-12
(4)
A(L')
=a
(5)
Let
J(L)
(6)
J(L') = J(L).
Now let
equivalent to Proof.
g3(L') = a
-6 g3(L).
2 - 27g3(L)
(7.33:22); thus
A(L). 3 g2(L)/A(L);
L
and
X ' :C/L'.
Theorem.
and
3 A(L) :g2(L)
Recall also that
and let
L'
.
Recall (7.33:18) that
(31
and
= u-ns, (L)
Sn(L'
140 s6(L)
L
be any lattices in C ,
let
x i C/L
Then we have L
If XI)
L'
then
,
is equivalent to
then
J (L)= J (L')
L'
(resp. X
is
.
( 6 ) gives the non-parenthetical result.
Theorem 8.45, the parenthetical result is proved.
Using
The Elliptic Modular Function
169
Finally note that
Let
9.12
L
L
be a lattice in C .
is a discrete subset of iI:
Since
by definition
there exists
E
L* (EL - C O I )
such that (1)
Clearly
for
1
E
L*.
has the same property.
-wl
Among all elements of
(2)
1 1 1,
is minimal among all
lull
L-Zwl,
choose
so that
w2
Iw2)
is minimal. Let
(3)
T
Note that, without loss of generality,
:w2/w1.
we may assume that Lemma.
R
Proof.
Let
Since
T EIR.
subset of
€6,
) is a basis of L. 1 2 L' :Zol+Zw2. Assume, for a moment, that
5 (w w
L
is a discrete subset of C ,
mul.
By the choice of
TPE.
absurd, proving that
R
T
is a basis of
L'.
L'
is a discrete
wl(l), L' = Zol; which is
Hence
L'
is a lattice in C
and
Assume, for a moment, that there exists
mcL-L'. ( P ( f l ) + u 1 L'
(4)
is a partition of C ;
thus, there exists a unique m o E m - 1 ; then
for
x
and
x , y ~(0,l).
y
mo
in
E
P(R).
roll).
By (1) and ( 2 1 ,
1 E L' As
such that
and w 2 .
Since lull
mPL,
P(R) + 1 . Let
zIw21
L' = L ,
Hence
mo#O.
(Imol;
thus
mo
can-
whose vertices are
As a consequence m l - w1 + w 2 - m o
is absurd; proving that
E
a consequence mo=xwl+ywZ,
not lie in the (closed) triangle T 0, w,l
m
E
LnTI
proving the lemma.
which
170
Norman L. Alling
R of
satisfying (1) and ( 2 ) will be called a minimal basis
L.
If, in addition, it satisfies ( 3 ) it will be called a t minimal positive basis of L. Let L: Z + i Z ; then (1 i) (i - 1)
and
are both minimal positive bases of
a minimal positive basis then so also is of
Let
9.13
L
be a lattice in
be a minimal positive basis of T
Let
E & .
L' 5wi1L
minimal positive basis of a basis of
and
L
L'
Proof.
Since ( 9 . 1 2 : l and 2 ) hold for
1' 1 1;
T
-1
would be in
Re T = t 1 / 2
are equivalent
L'
and
a',
' 1
IT-11
]TI.
Were
would be
then
(1
T
-1/2 < ReT.
7 l ) t is also a normalized
L';
can always be chosen so that
1 / 2 < Re
Note:
is such
which is prohibited by ( 9 . 1 2 : 2 ) , proving that
minimal positive basis of
-
R'
T.
Similarly we can show that
T
is
-1/2 < R e 'I < 1/2.
then
thus
R'
.
and
If
then
clearly
wl=l;
12 I T [
T >1/2,
t)
will be called a normalized
The following hold for
Re T 1,
or
1).
Given any lattice
L
which has a basis
R'
in C =
it is equivalent t (1 T ) , for some
D.
Definition
9.20
(1)
For
then
LT
(2)
Let
T E
0,
let
and
elementary properties.
LT:
2 +TZ,
is a lattice in
and
and let QT
(1
QT
T)~;
is a positive basis of
J ( T ):J ( L T ) .
J ( T ) is the elliptic modular function, one of the most inter-
esting €unctions in analysis.
For
9.21
(: :)
ME
E
GL2(c)
and
z E C,
recall
(3.13 :2) that
(1)
h(M) ( z ) z (az+b)/(cz+d) Let
Lemma.
(2)
J(T)
=
and let
Q' 5 M ' R T
Let
MI
5
t
(: .
z).
M E SL2 (Z);
LT.
5 (w'
(By (9.11:6),
then
M I is in SL~(Z). -1 I," :( u i ) LT, and let Clearly
') Let 1 w2 z o'/w' 2 1; then L" is a lattice in
Let
to
CQ
J(h(M) ( T I 1 .
Proof.
T'I
T
E C.
J(L") =J(LT).
which is equivalent Since
detM'(=1) > 0,
172
R'
Norman L. Alling
is positive (Theorem 6.22) )
L" = LTll. Clearly
that
hlSL2(Z)
{+I1 ; (4)
Q
and hence
proving the Lemma.
We have seen (3.14)
be called the modular group.
?I
and let
h (m)(T) ,
E
TI'
r :h(SL2 (2)) ,
Let
(3)
T" =
thus
;
is a homomorphism into
conQ
having kernel
thus
r
is a subgroup of S L (~z ) / { t 1 1
=
conQ
that is isomorphic to
P S L (2) ~ ,
the projective special linear group of rank 2
(2)
2.
may be rephrased as follows (5)
J ( T )= J ( g ( T ) )
thus
J
,
for all
T
E Q
and all
g
E
r;
is invariant under the action of the modular group on
63.
9.22
Let
Lemma.
n
E
N,
with
n13;
then
T
E
Q > -
s
n (L7 )
is analytic. Recall ( 7 . 3 3 : 5 ) that
Proof.
(a,b)# (0,O) ,
such that
a,b E 2 (1)
T
Let
E > O
E
p, >-
(atb?)-"
and let
QE:
1;
(GT)
5 8/Enkn'll
then clearly
is analytic. C T E Q : Im
notation of Theorem 7.31, h)~, (2)
sn(L) E CLEL* l-n. Given
T > E ~ .
Let
T E Q ~ .
and hence, by (7.31:6),
for all
k
E
N;
(3)
(4)
Since
Q = u
E>O
Using the
QE
we can combine (1), (3), and ( 4 ) to prove the Lemma.
173
The Elliptic Modular Function Since
g (L)= 140 s (L), we may use 3 6
g2 (L)= 60 s4 (L) and
the lemma to prove the following. Theorem.
analytic, for
t
gi (LT), A (LT
Q-->
E
j= 2
and
Let
t(z) s z + 1 ,
each
n E Z.
(2)
Let
then
h(T) = t,
(3)
T ~ =1
(o
n) 1
,
(4)
Let
S
(y
):-
(5)
(6)
1>
(o1
T-
S E
1
E
thus
SL2(Z),
t
1
(9.21:3).
E
T'.
nE
tn(z) = z
+ n,
for
z.
and let 2 S =-If
8
and
Clearly
Clearly
for all
Let
r
4.
of
SL~(Z);
for all
z
s : h(S).
and
s
€4,
and
be real, with
E
r. s
2 =l.
r>O
and
O < 8
It would
and
T'
IT^
but
above (3) is an autornorphism of order 2 of SL2(Z).
E
T ~ E D , and
are congruent mod
Note, in passing, that A
an automorphism of
D
are congruent mod
< Re T < 0,
~ / 2< 6 < 2 1 ~ / 3 ; T~
t,
such that
SL2(Z)
By Lemma 9 . 1 3 ,
(w1)-'L.
r.
E
is a normalized minimal positive basisl
(w1)-'O
let
-r0
I'.
there exists A
congruent mod T = ei6 , with Clearly
11.
L' - L T l l let
T = h(B) (T') ;
Re T = - 1 / 2
T'
1, or
be a minimal positive basis of
then
,
IT^ 2
and
1 ~ >1
and
A' :B ,
that induces
GL2(C)
of course
as defined
-
be pleasant
if it was not necessary to employ this automorphism. This could and writing a normalized be accomplished by letting T E w1/w2 basis as ( T 1)t The reason we have not done this is that it
.
is at variance with most of the classical literature. We have continued to use the classical conventions
-
in most cases
-
that e.g., [36] can be used virtually without modification.
so
The E l l i p t i c Modular F u n c t i o n
rr
Clearly
Let
(3)
then
A -
+
C T ~
ri
implies
f
cx
2
+
g = s,
-bE
(d-alx
(9.23);
rp
thus
Let
s
E
If
c#O
may assume
c#O.
a =d
thus
i s then a root
i and
a=O.
b = -c.
Hence
Since
A=*S,
p :e iT/3
and
qL
i s t h e f i x e d p o i n t of
rP.
are in
g c
-
f ( x ) Zcx thus
{lj; 2
+
then
l=ad-bc,
+
c#O.
Note t h a t
(d-a)x-bEZ[x];
a-d=c=-b.
w i t h o u t l o s s of g e n e r a l i t y
Since
a 2 - ac
rP -
then
Since
-
that
p
and
h(-A) = g , c>O;
thus
w e see t h a t
(c2-1) = 0 .
Solving ( 6 ) , u s i n g t h e q u a d r a t i c formula g i v e s
Since
a
then
ri.
c = * l and
that
l,q,
f ( x ) = 0;
satisfy
(6)
rT:
= {ilq,q2}.
p2-p+l=O.
a >d.
!g E
establishing ( 4 ) .
Indeed, l e t
-p
g=1.
then
Z[xl :
c#O,
Recall (9.23:20)
(5)
and hence
Clearly
r i - 11);
gc
and s i n c e
det A = l
q
h(A)
= 0.
A = + I,
Q.
such t h a t
= {~,sI.
f (x)
and so
-b
(d-a)r
Indeed, l e t of
SL2 ( 2 )
E
i s quadratic over
(4)
It i s called the isotropy
T.
c=O
Clearly T
at (: :)
r
subgroup of (2)
r.
i s a subgroup o f
177
i s real
c2 < 1: i . e .
I
c = 1.
Hence
b = -1 I
and
we
178
Norman L. Alling
ad=0.
If
a=O
then d = - 1 and g = q . 2 g=q establishing (5).
a = l ; hence
9.26
then
I
Since
is analytic on
J
variant under the action of
0
d=O,
~f
is the strip
TI
and since it is in-
it is periodic of period
(1) (6.50:l).
So,+m
Q
Thus
J(T)
1.
has a Fourier
series expansion
J(.r)
which converges to
uniformly on compacta of
Q.
(See
6.5 for details.) Let (3)
then
.
w - e2 n i ~I
a wn
m
ln=-m
converges uniformly on compacta to the
analytic function J(w), AOl1 Z { z Clearly
E
0
r
is
is analytic.
Norman L . A l l i n g
180
-
Lemma.
J(r(?)= ) J ( T ) , for a l l
Proof.
Let
T
J(r(r))=J(r(LT)).
for a l l
(3)
n> 2.
r c Q .
@. C l e a r l y r ( LT ) = Lr ( T ) ’ L e t L S L . R e c a l l (7.33:5) T
thus
E
-r -2n ~( L ) ) =~CLeL* ( (-L)
~
Clearly
g j ( r ( L ) )= g j ( L ) ,
that
j =2
for
and
3,
and
A(r(L))
p r o v i n g t h e Lemma.
i s r e a l - v a l u e d on
C o r o l l a r y 1. J
+
(4)
{n/2
J(T)
= J (T) ;
iy: y > 0 1 ,
p e r i o d 1, then
and l e t
proving t h a t
+
J(n/2
+
J(1/2
y > 0
Let
Proof.
proving t h a t
Since
Since
n
for a l l
E
22.
J
r ( T ) =
r,
i s p e r i o d i c of
r ’ -= 1 / 2
Let
+
iy;
+ iy) = J(r(1/2 + i y ) ) = J ( 1 / 2 + iy)
iy) = J(-1/2
+
T !i y .
J ( r ) EIR.
i y ) clR
J(1/2
n c Z.
f o r each
iy) E D .
Since
i s p e r i o d i c of p e r i o d
J
1, ( 4 ) h o l d s . Recall t h a t
i s r e a l - v a l u e d on
C o r o l l a r y 2. J
(5)
C1(0) n
Q.
Proof.
Let
Clearly
r s ( T ) = ~ .
i s t h e u n i t c i r c l e w i t h c e n t e r 0.
C1(0)
T
E
C1(0) n 0 ;
Since
J
J ( T )= J ( r s ( T ) ) = J ( s ( T ) = )
‘r = e i e ,
with
is i n v a r i a n t under
m, p r o v i n g
a (clcD).
i s r e a l - v a l u e d on
4.
J
i s r e a l - v a l u e d on e v e r y image of
r.
g3(Li) = 0 ;
thus
J ( i ) = 1.
T.
the Corollary.
J
( 5 ) under
0 < 0
0.
Note that
s(iy) =i/y.
Since
J
is invariant
s,
(4)
J(iy) = J(i/y) ,
(5)
thus
Y E
Y E
[ I r a ) ->
(0,1] ->
J(iy)
analytic map onto (7)
J(z) -1
(8)
Hence
y > 0;
for all
J(iy)
ing analytic map onto (6)
34
4
0
Note that
+ ti/2
is a monotone strictly increas-
[ I r a ),
is a monotone strictly decreasing [l,m).
From (2 we see that
has a double zero at
y=l
and
i.
is a relative minimum of
J(iy).
Norman L. Alling
184
Now consider relection across
9.32
of radius
1
1.
and center
C1(l),
the circle
This reflection is accomplished
by (1)
€ C ->
2
Under the map maps to 1/2
+
(1/2
Z / ( E - 1). C1(0),
+ IRi) ,
i/2 (3)
{1/2
(2)
+
Y E
i
maps to
+
1/2
i/2,
1
and center
and
p2
Since we know about the behaviour of
C1(0) n s) between on
the circle of radius
i
and
J(1/2
+
iy)
on J
is an unbounded, strictly
decreasing analytic map onto We know (9.28:7) that
J
we have the following:
ti: t > 31'2/21r
[1/2,m) ->
to
and about the behaviour of
p,
0,
J(p) = O .
J(z)
has a triple zero at
(3)
J' ( p )
(-m,ll.
From (9.31:2) it is clear that thus
p;
= 0 =J"(p).
Now consider reflection about
ClI2(1/2).
This can be
effected by z E C ->
(4) Let
- 1).
y > 0, and note that under (4)
(5) Since (6)
Z/(2'i
1/2 J
+
yi/2 ->
1/2
+
i/2y.
is real-valued on
J(1/2
+
yi/2)
=
J(1/2
ClI2(1/2)
+
,
(9.31:3) ,
i/2y).
From (2) we see that
(7)
Y E (0,1] ->
J(1/2
+
iy/2)
increasing analytic map onto (8)
Further, 1 for
is an unbounded strictly (-all].
is the absolute maximum of
y > 0, and
J'(1/2
+
i/2) = O .
J(1/2
+
iy/2),
The Elliptic Modular Function
Having raised questions about isotropy subgroups
9.33
r
of (1)
in 5 9 . 2 5 , Let
and
T
I
let us give a complete analysis of them.
Indeed, for
4
be in
and let r T I = g T T-1 g
T'
f E TT,
g-lf ' q :f;
q
r
E
such that
.
then
T ' = ~ ( T ) ;
and let
185
gfq-l E F T l .
then
then
f E TT
and
f'
Let
qfq-'=f',
E
rT,
establishing
(1).
Let
(2)
and let
T
Do
be the interior of
D0
E
.
rT= i l l .
(3)
Indeed, suppose
for a moment
There exists
is not 1. Do
that is transformed to
T~
E
U
Let
(4)
T
aD- {i,pj; then
E
Indeed, assume E
TT
-
{l}.
Since
real locus of a sub arc
orientation of
g,
aD
A.
But
J
aD,
U
about
T
that in
# T ~ ;
but
establishing (3). T T = {l}
that there exists
is invariant under
into
rT
where -r1
is mapped to itself under
J
of
A
by
for a moment
J
gc
there exists
in an open set
T~
this contradicts Corollary 9 . 2 7 ,
g
D,
fixing
T,
is monotone on
r
(9.21:2),
g.
Hence
the g
maps
and reversing the
A;
which is absurd,
establishing ( 4 ) . Thus we have the following, using ( 9 . 2 5 : 4 Theorem. p.
T i = {l,s}
9.40
Let
For and
TED,
TT={l}
T p = {l,q,q
L
and 5 ) .
if and only if
Tfi
or
1.
M o d u l a r functions
f
be a monomorphic function on
0
that is invariant
186
Norman L. Alling
r
under the action of
8.
on
there exist constants c
(1)
for all Let
F(r)
y
im(z)
is a modular function if
f and
k
such that
greater than some
If
(2)
yo EIR.
be the set of all modular functions. Clearly such an
.
F(r)
element in
C
(2)
is a simple transcendental extension of @ .
C(J)
(J) is in
Let
Proof.
f E ~ ( r ) . f maps
invariant under the action of D,
Further, clearly
F ( T ) = C (J).
Theorem.
on
r,
a fundamental domain for
function F
on C
such that
8
and since J
r,
.
f = F (J)
F
thus
f
is
is injective
(1) and Theorem 9.26,
Theorem 8.30,
Q: ( z ) ;
Since
there exists a monomorphic
extends to an analytic map of E
Z.
into
imply that F
into
C
1.
By
f E C (J), proving the Theorem.
i n v e r s i o n problem
9.50
Let
-< cekY
a
and
b
be in C
such that
2 a3 - 27b # 0.
The
Weierstrass inversion problem is the following: (1)
Find a lattice
L
in Q:
such that
g2(L) = a
and
93 (L)= b* Theorem.
The Weierstrass inversion problem always has
a solution. Proof.
Assume first that
afofb.
If
L
satisfies (1)
then (2)
J(L)
=
a3/(a3-27b2)
Conversely, assume that see that (1) holds. fies (2).
and L
g2 (L)/g2 (L) = a/b.
satisfies (2); then one can easily
Thus it suffices to find
By Theorem 9.27 there exists
T E
D
L
which satis-
such that
The E l l i p t i c Modular F u n c t i o n
3
3
21 .
J ( T )= a / ( a -27b
(9.11:6)
J ( L ) = J ( - r ) . By ( 9 . 1 1 : 3 )
2
cx g 2 ( L T ) / g 3 ( L T ) . C l e a r l y
let
T:
then
p;
# 0, g3 (LT) # 0.
A(LT)
g2(L,) that
a3
-
i;
T
# 0.
then
g 2 ( L ) = cx
g2(L) = a r
a # 0 # b.
(L ) ;
3 . r L a s t l y assume t h a t
(9.28:6).
g 2 ( L T ),
Since
g 3 ( L ) = a-6g
g3(L) = b .
If
Since
thus
ci
b=O.
A(LT) # O r
thus there exists
cx E C
such
p r o v i n g t h e Theorem.
Now assume t h a t a and b a r e i n IR s u c h t h a t 2 27b # O . Assume f i r s t t h a t a # O # b ; t h e n (1) and ( 2 ) a r e
again equivalent.
If
A
unique
J ( r ) = a3/(a3-27b 2 )
such t h a t (3)
g2(L)/g3(L) =
g 2 ( L T )= 0 ( 9 . 2 8 : 7 ) .
g 3 ( L T )= b -4
t h e n by
may be c h o s e n s u c h t h a t
ct
Note t h a t
may be c h o s e n so t h a t Let
,
L : aLT;
p r o v i n g t h e theorem i n c a s e
g2 (L)/g3 (L) = a/b, a =0
and l e t
a€@
Let
187
may b e found i n
.r
aD+ ( 9 . 2 8 : 8 )
.
g 2 ( L T ) / g 3 ( L T ) and
have t h e same ( r e s p . o p p o s i t e )
a/b
s i g n , t h e r e e x i s t s a unique
ct
> 0
such t h a t
ctTT
( r e s p . icxT ) s a t i s f i e s ( 1 ) . T
Assume now t h a t
(4)
If
a=O;
g3(LT)
and
then
b
t h e r e e x i s t s unique
e
in/6
cxLT)
(5)
If
b = 0.
A(LT) # O r
g 2 ( L T ) and
-
b#O.
Let
T:
p.
have t h e same ( r e s p . o p p o s i t e ) s i g n , cx> 0
such t h a t
cxLT
(resp.
Let
.r=i;
then
g3(L,) = b .
g2(L,) # O . a
t h e r e e x i s t s unique
ein’4cx~.r)
of cours e
s a t i s f i e s (1).
L a s t l y assume t h a t Since
-
have t h e same ( r e s p . o p p o s i t e ) s i g n ,
a >0
s a t i s f i e s (I),.
such t h a t
aLT
(resp.
This Page Intentionally Left Blank
CHAPTER 10
ALGEBRAIC FUNCTION FIELDS
Definitions
10.10
Let that
L
F(L)
and
Introduction
be a lattice in
@.
We have seen in Chapter 7
is a field extension of
as follows. p
which can be described
@
is transcendental over
finite algebraic extension of
@(?).
C
and
F(L)
is a
This will serve as a model
for the definition to follow. Let
K
be a subfield of a field
F.
KcF
will be called
an algebraic function field of one variable, or merely an algebraic function field, if there exists over
K,
such that
F
x
E
F,
transcendental
is an algebraic extension of
K(x)
of
finite degree. Example.
@ c
F(L)
is an algebraic function field.
The theory of algebraic function fields is a large one. We will give only a very brief sketch of it in this chapter, giving frequent references to the literature.
Hopefully we will
write enough to motivate what has been done when
K
is IR
or
C, and thus make Part I11 of this monograph more accessible to the reader not knowledgeable in these matters. The standard reference to this subject is Chevalley's concise and extensive Introduction to the theory of algebraic functions of one variable [16]. A more classical reference to the state of the art c. 1900 is Hensel and Landsberg's Theorie 189
Norman L. Alling
190
der - algebraischen Funktionen einer Veriabeln 1341.
Let
10.11
(1)
Let
E
XI
KcF
F
be an algebraic function field (10.10).
be transcendental over
K;
Indeed, since the transcendence degree of definition, 1, x
is algebraic over
[F:K(x)], we see that
[F:K(x,x')]
then [F:K(x')]
(1)
Let
fc F
gives rise to a map
?(0) E
C
(10.12:4).
have the weak topology making (1) continuous, for a1 L
X
f E F. (2)
Each
Chevalley [16, p.133 ff.] has proved the following. X
is a compact space.
As remarked before ( 1 0 . 2 0 : 7 ) , thus there exists
2;
fE F
the value group of
such that
0
is
vO(f) = 1.
Chevalley has also shown that (3)
f
is a local uniformizer at
0; thus X
is a compact
surface. Finally Chevalley has shown that the Atlas so defined is analytic ( 8 . 2 0 ) (4)
X
;
thus
becomes a compact Riemann surface X
is a meromorphic function on
X;
and
such that each
f E F ->
2
E
F(X)
is
a surjective @-isomorphism. Remark.
structing X
In the author's opinion,Chevalley's method of confrom C c F
is a great improvement over the "cut
and paste method" described in 58.13.
4 theorem
10.40
Let
C
of
coeguivalence
be the category whose objects are complex alge-
Algebraic Function Fields
201
braic function fieldsand whose morphisms are @-linear isomorphisms of one such object into another.
S
Let
be the category
whose objects are compact (connected) Riemann surfaces and whose morphisms are analytic surjections of one such object onto (See,e.g., 1501 for the revelent definitions in cate-
another.
gory theory.) let
f
Let
@
cF
and let
of generality, we may take @
@ cF
and
cF
of the morphism
f.
Let
into
be objects in
F.
[F : F].
Riem F
5
X
and
@
n
(Riem f) ( ? ) E
then
Riem f
@
C F is a finite
-
,..
X :Riem F ;
c
then,
S.
a
Given
let
E
2 n F;
Riem@
Lemma 1.
F
Since
are compact Riemann surfaces:
is an analytic map of
a:
-
F.
is called the degree
and let
each is an object in the category
(1)
and
C
Then, without l o s s
to be a subfield of
n: X
Let
as we noted in 510.30, X
,
F
-
cF
are algebraic function fields
algebraic extension.
i.e.
F
be a @-isomorphism of
@
2 onto
X.
Hence
is a contravariant function of
c
into
S. Y -> a
Now let object in
Let
S.
Y.
tions on
for details.) onto
g
is an
be a triple of object, morphism, and
F(Y)
be the field of a11 meromorphic func-
Using the fundamental existence theorem on Riemann
surfaces, F(Y),
Y
Y
is a proper extension of
Let
g
E
F(Y)
then
-@;
(2)
and
is an analytic map of
It may be shown that there exists
C.
n - to - 1 map of
Y
onto
Z
Then it is not difficult to show that F(Y)
f
(See e.g. , [ 5 8 1
@.
F(V)
are objects in
F(a) (h) ha,
to show that
for all
h
6
c. F(Y)
nc N
such that
(counting rami€ication). [F(Y):
@
(g)]
=n;
thus
Let
.
Then it is not difficult
Norman L. Alling
202
Theorem of c o e q u i v a l e n c e .
Riem
and
Q:
s
is a contravariant functor of
F
Lemma 2 .
into
c.
The contravariant functions
establish a coequivalence of the categories
F
c
S.
and
(See [5, pp.95-1041 for details.) f is a triple of object, morphism, and Thus if F1 > F2
C;
object in (3)
F1
L.
If
X 2 ->
in
S,
then
@ F(Rie% F1)
a
F (Riem@f )
> F(Riemc F 2 ) = @ F 2 .
is a triple of object, morphism, and object
X1
then Rie%F (a)
(4)
> Riem F(X1) =
X 2 = Riem F ( X 2 )
c
c
Let
Theorem.
x :C/L
and
let F '
X':@/L'.
F(L')
L
and Let
L'
F
xl.
be lattices in
@.
Let
F(L) (or equivalently F ( X ) )
(or equivalently F ( X ' ) ) .
and
The following are
equivalent: (i)
L
and
L'
are equivalent lattices ( 6 . 2 3 ) .
(ii) X
and
X'
are analytically equivalent.
(iii) F
and
F'
are @-isomorphic.
(iv) J (L) = J (L') Proof.
.
By Theorem 8.45, (i) and (ii) are equivalent.
Theorem 10.40, (ii) and (iii) are equivalent.
By
By Theorem 9.29,
(ii) and (iv) are equivalent; proving the Theorem.
The Riemann-Roch Theorem
10.50
Let
X
be a compact Riemann surface, and let
the underlying topological space.
Since
X
X
denote
is a compact
orientable surface,it is homeomorphic to a sphere to which
g
Algebraic Function Fields handles have been adjoined. X
called its genus.
g
203
is a topological invariant of
(See e.g., [51] as a general reference on
the topology of surfaces.)
Let
a
be a divisor on
X.
(See
58.60 for a definition.)
(1)
Let
L(a)
then
L(a)
is a complex subspace of
(2)
L(a)
is of finite complex dimension
Riemann's
(3)
L(a)1'
(4)
Let
{f
E
F(X): (f) >a]; F(X)
. L(a).
Theorem
- g - deg(a) .
i(a),
the index - of specialty of
be the non-negative integer
a,
be defined to
-1 + L (a) + deg (a) + g;
then we have
the following. The Riemann-Roch
Theorem.
For all
a
E
divX,
the follow-
ing holds :
(5)
O=l-.l(a) -deg(a) -g+i(a). The Serre D u a l i t y Theorem.
For all
aEdivX
c U (X) (-a), where
(6)
i (a) = dim
(7)
U(X) (-a) :(6
(Note: D(X)
E
D(X):
(6) > -a}.
was defined in 1 8 . 2 3 . )
The literature on the Riemann-Roch and the Serre Duality Theorems is extensive.
Nevertheless, Gunning's excellent
Lectures 0" Riemann surface [29] would serve to give complete proofs of these beautiful and important results.
10.51
The Riemann-Roch theorem can be stated and proved
for any algebraic function field 113.)
KcF.
In so doing a non-negative integer
the genus of
KcF.
(See e . g . , g
[16, Chapter
emerges, called
One of the great accomplishments of contem-
204
Norman L. Alling
porary algebraic geometry is that it has been able to carry over much algebraic and analytic geometry, that was first discovered and investigated over
@,
to arbitrary ground fields.
PART I11 REAL ELLIPTIC CURVES
This Page Intentionally Left Blank
CHAPTER 11
REAL ALGEBRAIC FUNCTION FIELDS AND COMPACT KLEIN SURFACES
Real
11.10 IRc K
Let
algebraic function
be an a l g e b r a i c extension,
a l g e b r a i c a l l y closed; then to
fields.
-
of course
f o r which
-
K
is
K
is
IR-isomorphic
@.
(1) T h e r e e x i s t e x a c t l y two IR-isomorphisms, onto
K
R
Let
Further
@.
CY
=
j
CY 1-j'
for
c1
0
j = 0
and
all
and
1.
d e n o t e t h e c a t e g o r y whose o b j e c t s a r e r e a l a l g e b r a i c
f u n c t i o n f i e l d s ( 1 0 . 1 1 ) and whose morphisms a r e
c
F
be an o b j e c t i n t h e c a t e g o r y
C
IR c
CI
Since
function f i e l d s (10.40).
R.
@ c
R.
Let
F
+
lRcE
R
be an o b j e c t i n
Let
of complex a l g e b r a i c
IR c F i s a n o b j e c t i n
is then a covariant functor
IRc F
in-
IR-linear
j e c t i v e homomorphisms o f one s u c h o b j e c t i n t o a n o t h e r .
C
of
and l e t
5
of
C
into
b e i t s f i e l d of
constants (10.11). (2)
If
-
E i t h e r ( i ) IR i s
IR,
o r (ii) it i s
( i i ) h o l d s t h e n (1)
IR
h a s two d i s t i n c t IR-isomorphisms
IR-isomorphic t o
@.
-
onto
@.
Using e i t h e r o f t h e s e w e c a n make
i n t o an o b j e c t i n
c.
( W e w i l l see l a t e r t h a t t h e s e two o b j e c t s
aO
and
a1
need n o t b e e q u i v a l e n t i n Assume now t h a t
f i e l d of
x2
+
1
C.)
( i )h o l d s i n ( 2 ) ; t h e n
i r r e d u c i b l e polynomial i n over
E;
IRc E
E[x]. then
207
Let IRc F
F
x2 + 1
is an
be t h e s p l i t t i n g
i s an o b j e c t i n
R
208
Norman L . A l l i n g
-
whose f i e l d o f c o n s t a n t s s a t i s f i e s ( i i )o f
IRc F
in
IR
F.
Let
(2).
-
renders
denote a r o o t of
and
t h a t takes
F
h a s a genus
g.
-
of
cil
+
i
to
with
IR
@.
x2
C
This o b j e c t
IR c E .
i s t h e f i x e d f i e l d of
E
E
IRc
cxo
Hence
@.
and w i l l be c a l l e d
@ c F
t h e complexification of
Clearly
i
Let
i n t o an equivalnet o b j e c t i n
IRc F
w i l l be d e n o t e d by (3)
IR-isomorphic t o
b e t h e E-automorphism o f
0
Thus e a c h IR- isomorphism
-1.
is
A s noted i n ( 1 0 . 5 0 )
0.
,
C h e v a l l e y [16, p . 991 h a s p r o v e d t h e
following: Theorem.
i s a l s o o f genus
@ c F
g.
11.11
constants i s
is
tion
c
c
whose f i e l d o f
Its complexifica-
IR, and whose g e n u s i s z e r o . ~ ( x ) .
Example 2 .
IRc
@ ( x ) i s an o b j e c t i n
-
IR i s IR-isomorphic t o
constants
R
is an object i n
IRc IR(x)
Example 1 .
R
whose f i e l d o f
and whose g e n u s i s
@,
zero. The f o l l o w i n g i s a more i n t e r e s t i n g e x a m p l e , which may b e found e . g . ,
i n [16, p . 231.
of
F
F :@ ( z ) . L e t
Let
Example 3 .
b e t h e IR-isomorphism
0
t h a t e x t e n d s complex c o n j u g a t i o n and maps
z
to
-l/z.
( T h a t s u c h a n automorphism e x i s t s f o l l o w s from t h e f a c t t h a t
-l/z
generates
t r a c e of
f,
(1) L e t
x :T ( z ) / 2
then
be
over
F
f
x = ( z - 1/2)/2
+
C.)
For
f
E
F
a(f). and l e t and
y 5 T(iz)/2;
y = i(z+l/z)/2.
let
Tr(f), the
1
209
Real Algebraic Function Fields
(2)
Hence
Let
E
IRc
E;
thus
E
-1.
=
R.
is in
IRc E
Since
0.
Clearly
x + y/i
Since
z,
=
is
CJ
x
R-linear,
and
y
E;
are in
IR(x,y) (i) = F ;
thus
IR(x,y)
=
R
is an object in
IRc E
= @(z).
(4)
+ y2
be the fixed field of
Clearly CC
2
IR(x,y) c E.
thus (3)
x
whose complexification is
Using Theorem 11.10, we see that
E
IRc
is of genus
0.
It has been shown [ 6 1 that (5)
IRc E
If
R
is an object in
of genus
0, then it is
equivalent to the object considered in one of the three examples above.
Now let us consider examples of objects in
11.12.
Let
Example 1 .
- 27g32
2 and let
y
in
R
q2
Let
be in
x
g2x
=
of genus
w
such that
v
g3.
then by Theorem 11.10, IRc E
Let
be in =
such that
is an object
1.
E x a m p l e 2.
u, v, and
-
IR such that
be transcendental over IR
be algebraic over IR(x)
IR(x,y);
E
and
g1
is non-zero.
(1) y2 = 4x3
Let
of
1.
genus
A 'g3
R
w
x
IR.
be transcendental over
{O,l};
implies
and let k < 1.
(2)
Lu ,v w (x,k) : (-l)'(l-(-l)vx2)
Let
y
be algebraic over
(3)
y2
=
Let
E : IR(x,y)
I
IR(x) ,
k
E
IR,
with
Let 0 < k < 1,
Following (3.21:10), let w 2 2
( l - ( - l )k x 1 .
and let
LUIVIW (x,k).
.
By Theorem 11.10, I R c E
is an object in
2 10
R
Norman L . A l l i n g
1.
of genus
I t w i l l b e shown i n t h i s monograph t h a t Examples 1 and 2
R
g e n e r a t e a l l examples of o b j e c t s i n
1.
of genus
Further,
e l e m e n t a r y c r i t e r i a w i l l b e g i v e n which w i l l a l l o w u s t o
R.
d e t e r m i n e when any two s u c h o b j e c t s a r e e q u i v a l e n t i n
11 . 2 0 .
Klein Surfaces
R,
be a n o b j e c t i n
IRc E
Let
and l e t
IR
denote i t s
f i e l d of c o n s t a n t s ( 1 0 . 1 1 ) .
(1) L e t Given
(I
IR
let
X,
E
By ( 1 0 . 2 0 : 4 ) ,
(2)
Thus
(3)
Let
Let
f
E
ax
(4) (I E
Let
x w
is e i t h e r :
(I E
X:
and l e t
(10.12:4);
thus X
denote i t s r e s i d u e c l a s s f i e l d ( 1 0 . 1 2 ) .
EO
is a f i n i t e a l g e b r a i c extension of
El)
EO
E
(10.11) :6)
Riem E
X
If
X.
f If
IR.
E
E
f
0) t h e n
?((I) E
li((I) I
,! 0 l e t
I ; ~ C O )I
continuous, f o r a l l
89 f f ]
.
X
f
E
f
m.
E.
i s a compact c o n n e c t e d s p a c e .
L e t u s s k e t c h a p r o o f o f t h i s t h e o r e m , which
w i l l prove u s e f u l .
IR
and
C
X
is an o b j e c t i n
E
X
i s i n t e g r a l l y closed (10.12),
C.
Let
choose an
be i d e n t i f i e d using it; then
c
are identical.
in itself
i s I R - i s o m o r p h i c t o C,
If
i s o m o r p h i s m and l e t
and Y
EO
b e g i v e n t h e w e a k e s t t o p o l o u y making
Theorem [6, p .
(I
C.
IR}.
EO =
(I E
IZ(0) I
or i s IR-isomorphic t o
IR
IR:
Y
f
Since each
RiemlRE. -
IRc
thus, as sets,
(I;
As n o t e d i n (10.30),
Y
can be given
t h e s t r u c t u r e o f a compact Riemann s u r f a c e i n a v e r y n a t u r a l
X
Real Algebraic Function Fields way.
ing that
Y
and
is identical, show-
is a compact connected surface (without boundary).
X
-
IR = IR.
Assume now that
of
X
Clearly the topology on
211
Let
C
be the complexification
Riemann surface structure
Riem F be endowed with its C Y (10.30). Let u be the non-
trivial E-automorphism of
F
E (11.10:3) and let
IRc
a map
u*
Y
of
u*
that
Y
F
c
onto
f
of order
Y
is a homeomorphism of
Y
onto
X
to see that
Y.
Indeed, one
i.e., that it is
(See, e.g., 1 6 , p. 1 ff].)
2/32.
induces
It is easy to see
2.
easily shows that it is anti-analytic: annihilated by
u
(11.10). Clearly
It is easy
is homeomorphic to the quotient space
Y/u*;
completing our sketch of the proof of the Theorem.
Returning to the examples of $11.11, it is easily
11.21.
seen (for example by consulting Chapter 10)
that
(1) RiemRC (x) is homeomorphic to the Riemann sphere
C;
and that RiemnIR(x)
(2)
(where C+
{a
5
Example
3
is homeomorphic to
+ bi: a,b
(§11.111.
cendental over
IR
is
C
E(i); thus
RiemC@(z) C(z)
takes Thus
and
Let 2
E Z IR(x,y) where
+ y2
.=
u*
C*
to
-
l/h
E
C*,
is fixed point free.
is then homeomorphic to
x
is transF
is the complexification of
c C(z)
is homeomorphic to
E
I:,
-1. Recall that
C.
of
and which permeates The quotient space
RiemnE.
C(z)
IRc E.
The E-automorphism u
induces an anti-analytic involution o*
X
in
{m}
> 01). IR, b -
E
x
C+ u
X
of
I: which 0
and
C/u* : X
is easily seen to be
m
212
Norman L. Alling
homeomorphic to the real projective plane. orientable and has no boundary.
Thus
X
This suggests that
is non-
ax
(11.20:3) should be empty.
It should here be remarked that all of this is worked out in greater detail in [ 6 ] .
Y : Riem IRIR(x).
Since
IR(x) c IR(x,y) O
there is a map which takes
ax,
0'
For
0 to be in
M',
the maximal ideal of
r
IR.
E
Let
pO
then we see that
0
of
Thus
ax
To see why
E
X
4,
let
is a morphism in
0' : 0
to
must be in
o',
=
aY.
n
If
IR(x,y) E Y. x
E
0' then
(x-r)U', for some
must be
be the place associated with pO(x) = r.
R,
0
(10.12);
In the residue class field
EO
(10.12) we see that
cannot be the field
Eo
0' then
is not in
l/x
is.
IR, and hence must be Thus
0'
=
it is the point at infinity in Y. Then equation x 2 + y2 = -1 then gives us (4) 1 + (y/x)
Applying
po
2
= -l/x
2
@.
If
-
IR(l/x)
i.e., (l/x) * po(l/x) = 0. The
.
to (4) gives us
1 + (Po(Y/x))2 = 0;
(5)
thus
EO
cannot be the field
IR, and hence must be
@.
We
have thus shown that (6)
ax
=
This example was cardinal in motivating the research of Newcomb Greenleaf and the author which lead to [ 5 1 and [ 6 ] . It became clear that if X : Riem E
IR
IRc E
x
is an object in
R,
then
might be orientable or non-orientable; and that
213
Real A l g e b r a i c Function F i e l d s
ax
I n [ 6 ] w e showed t h a t
m i g h t o r m i g h t n o t b e empty.
(7)
X
was a l w a y s a c o n n e c t e d s u r f a c e w i t h ( p o s s i b l y empty)
ax,
boundary and
t h a t any s u c h s u r f a c e c o u l d a r i s e i n t h i s way.
(8)
One o f t h e main t h e o r e m s i n [ 6 1 was t o p u t a s t r u c t u r e s u c h t h a t i t s s e t o f "meromorphic" f u n c t i o n s would b e
on
X
E.
The s t r u c t u r e w e p u t on
ture.
we called a dianalytic struc-
X
T h i s i d e a g o e s back i n e s s e n c e t o K l e i n ' s 1882 monograph
[ 3 9 ] , f o r i n t h e c l o s i n g p a g e s of i t h e c o n s i d e r s "Riemann surfaces" t h a t a r e non-orientable.
S c h i f f e r and S p e n c e r [ 5 6 ]
c a r r y o u t t h i s program i n a c o n t e m p o r a r y way.
The i n t e r e s t s
o f G r e e n l e a f and t h e a u t h o r w e r e more i n c l i n e d t o a l g e b r a i c g e o m e t r y t h a n S c h i f f e r and S p e n c e r seem
t o have been; t h u s
o u r e m p h a s i s and o u r r e s u l t s w e r e i n many e s s e n t i a l s q u i t e different.
11.22.
The d e t a i l s on K l e i n s u r f a c e s may b e found i n [ 6 ] .
I n C h a p t e r 1, e n t i t l e d K l e i n S u r f a c e s , a n a l y t i c p r e l i m i n a r i e s may b e found i n $1.
A t h o r o u g h d e s c r i p t i o n of what a d i a n a l y t i c
s t r u c t u r e i s may b e found i n 5 2 . "functions".
5 3 d e a l s w i t h meromorphic
This occupies 1 6 pages o f t e x t and c o n s t i t u t e s
the kernal of the ideas.
less e s s e n t i a l .
§§4-8 a r e u s e f u l b u t a r e a l i t t l e
$ 9 on s u r f a c e s o f g e n u s 0 and 1 i s e s s e n t i a l
f o r t h e reader t o understand.
I n d e e d , t h i s monograph may b e
viewed a s a n e x t e n s i o n o f C h a p t e r 1, $ 9 o f
11.23.
[61.
The h i s t o r y o f K l e i n s u r f a c e s , a s i t was known by
Norman L. Allinq
214
Greenleaf and the author c. 1971 is sketched in [ 6 1 .
The two
additional references which have come to the author's attention
ijber symmetrische Periodicitatsmodulu der
in the intervening decade are Guido Weichold' s Riemann'sche Flachen und die
zugehorigen Abel'schen Normalintegrale erster Gettung [63], published in 1883.
See also
W. - D. Geyer's report given at
Oberwolfach in 1964, in which he uses Galois cohomology to obtain many of Weichold's results [281.
Symmetric Riemann s u r f a c e s
11.30.
X
Let
be a Riemann surface and let
analytic involution of
(1) Then
:Y
X/5
be an anti-
X;
is a symmetric Riemann surface.
(X,5)
(x,o)
Assume that
5
is a symmetric Riemann surface; then
is a Klein surface.
Y
is compact if and only if
X
is. Conversely, given a Klein surface orientable or for which X
[6, $1.61; then X/O
that
and
Y
to work on out on
X.
11.40.
Y.
let
which is non-
X
be its complex double
has an anti-analytic involution are dianalytically equivalent.
have preferred to work on differentials,
8Y # $,
Y
... .
X
o
such
Some authors
with symmetric functions,
The author and Newcomb Greenleaf preferred
Nevertheless, many of our proofs were carried
This point of view is continued in this monograph.
A_ t h e o r e m o f c o e q u i v a l e n c e
A theorem of coequivalence (similar to Theorem 10.40) for a category of real algebraic function fields and iR-linear
Real A l g e b r a i c F u n c t i o n F i e l d s
215
monomorphisms,and a c a t e g o r y o f compact K l e i n s u r f a c e s and K l e i n morphisms i s proved i n [ 6 , pp. 9 5 - 1 0 4 1 . one o f t h e main theorems o f t h a t monograph.
This i s
This Page Intentionally Left Blank
CHAPTER 1 2 THE SPECIES AND GEOMETRIC MODULI OF A REAL ELLIPTIC CURVE
The
12.10
Let lRcE Y
then
1
extended modular yroup
be an object in
R
of genus 1.
Let
Y Z
Rie%E;
is a compact Klein surface whose (algebraic) genus is
[4, p.241. D e f i n i t io n
(1)
lRcE,
Y,
or equivalently
will be called a
real elliptic
curve. Let lRcE
be a real elliptic curve with
As
Rie%E.
noted in 511.10 the constant field, (2)
-
is either (i) lR
R,
Assume that (ii) holds. phisms of % 1.
and
Let
and
(11.1O:l); then
a1
and
a:;
then a: c E
@.
be the twoIR-isomor-
-
aj =alejI for
j=O c1
1'
to
is an object, of genus 1, in
(10.40).
(3)
x 3.
5
Riem E
a:
is a complex algebraic curve of genus 1.
We have seen (10.30:4) that of genus 1.
a:
a0
Having chosen one of these isomorphisms, say
identify
C
onto a:
or (ii) it is Ill-isomorphic to
such that
X
is a compact Riemann surface j By Theorem 8.50 there exists a lattice L(j) in
x
and @/L(j) are analytically equivalent. In j Chapter 9 we saw that there exists a unique T E D (9.13:5), a j
fundamental domain of the elliptic modular function, such that (c/L( j)
and
@/LT
( 9 . 2 0 :1) are analytically equivalent.
j 217
By
218
Norman L. Alling
Theorem 7.42 (4)
E=@(P(',LT
)
i
j
P'(*iLT ) I * j
we have also seen (7.33:19) that
Now, of course, there is no reason to choose opposed to (6)
Since
ctl-j.
-
1'
gk(LTlej) =-,
=
as
j
we see that
'1-j
for
c1
k=2
and
3
and
1.
and
j=O
j
Recall (9.28:l) that a l s o that
gk(r(L)) =
r(z)
=-;,
v,
for all
z
E
fi.
Recall (9.28:3) By Theorem 9.27
for
k= 2
and
3.
for
j= O
and
1.
we see that :r(rj) - ~
(mod
r),
(7)
T
(8)
Let the extended modular group
T
lytic autohomeomorphisms of 9
generated by
~
Bibliographic note.
be the group of diana-
r
and
r.
[Klein-Fricke for example, considers
such maps [41, vol.1, p.196 ff.].
We have adopted the terminol-
ogy of DUVal [19, p.44 and p.2461.
Clearly
12.11
(1)
.. r=r
%
(2)
(9.28:l) is of order two, hence
u r r = r u rr; thus
T/l' =
and Clearly
r
r
Given
Thus
is a normal subgroup of
z2.
is the set of all elements in f c rr, there exist
f(z) = (aE+b)/(ci+d), (3)
r
D+
-r
a,b,c,dE 2
with
-T I
that are analytic. such that
ad-bc=-1.
(9.28:8) is a fundamental domain for
?.
Hence we have proved the following: Theorem.
Let l R c E
be a real elliptic curve, and let
219
The Species and Geometric Moduli Y
Rie
%E.
Assume that the field of constants
morphic to
i.e., that Y
@:
There exists a unique
T E
is IR-iso-
is orientable and has no boundary.
D+
Y
such that
and @/LT
are dian-
alytically equivalent.
12.20
Species
Let IRcE
be a real elliptic curve, let
assume that the field of constants assume that
or that
aY#B
out in (11.101, x L + l
of IRcE
splitting field of
x +1
of all elements of
F
and
is 1R:
i.e.,
is non-orientable.
Y
is irreducible in 2
Rie%E
Y E
over
E[xl.
Let 5
E.
As pointed
Let
F
be the
now denote the set
that are algebraic over IR:
i.e., the
field of constants of IRcF; then 5
(1)
is IR-isomorphic to
By Theorem 11.10, I R c F X:Rie%F.
is an object in
and @/LT
morphism
u*
alent to
x/a*.
of
equals
x (X) = 0;
thus
(2)
=o.
Let
X(Y)
s
of genus
are dianalytically equivalent.
the non-trivial E-automorphism of
x(X),
R
X
onto
X;
F.
then
Let
Let
0
such be
induces an anti-analytic
(5
Y
1.
+ TED
By Theorem 12.11, there exists a unique
X
that
@.
is dianalytically equiv-
It is easy to see that the Euler characteristic, 2x(Y).
Since
x
is of topological genus 1,
be the number of components of
Y.
Using the classifica-
tion of compact surfaces (see e.g., [51] for details), we find that the following holds: Theorem.
~f
Y
s=2,
Mobius strip, and if
s=
0
is an annulus, if it is a Klein bottle.
s=l
it is a
Norman L. Alling
220
The species of
(3)
is the integer s
Y
Y
Note that the species of
.
s( Y )
determines the homeomorphism type
of the underlying space Y.
Geometric moduli
12.30
Let I R c E
be a real elliptic curve and let
be the field of constants of IR c E .
Let
i.e. I that Y
is IR-isomorphic to C : that
aY =
a.
(1)
Now assume that z=IR: i.e., that orientable. Let
F
Let
X
(2)
y
of
a
E @ *
basis
Q
E
Letting
(4)
z =T
over
F; Y
then and
E
x. u*
is
X/u*
are
b
E
of
Z E C ,
P(~,T), the period parallelogram of the L .: LT implies
(6.20:6).
y
One easily sees that
(2) Z y ( z l )
and
z' = O
in ( 3 ) gives
and
z' = 0
in ( 3 ) gives
a+b
a? E L .
Note that
2 x +1
lifts to an anti-analytic automor-
forall
(mod L)
z=1
y
is non-
has an anti-analytic
a E L.
Letting (5)
and
(1 T)
z: z '
D+
which must have the form
Y(z).:ai+b,
where
(3)
Clearly
@,
Y
such that
dianalytically equivalent. Thus C/LT involution y.
E
be the geometric modulus of
T
ananti-analyticautomorphism of
phism
or
a Y f a
be the non-trivial E-automorphism of
CJ
T
Y.
of
be the splitting field of
and let X:Rie%F. Let
is orientable and
are dianalytically equivalent.
is the geometric modulus m(Y),
T
Rie%E.
Assume first that
By Theorem 12.11 there exists a unique
Y and C/LT
such that
Y
y2 = l x I
the identity map of
X.
(mod L). 5
b
(mod L);
hence
The Species and Geometric Moduli
(6)
y 2 (z)= a a z + a E + b , for all
-
aa=l;
la1 = l .
thus
y 2 = l X , y2(0)
Since (8)
is in
L;
thus
ag+bcL. ([6, (1.9.6)]).
Proposition.
either
Re(T)
is
is a basis of T
either
+; 0
or
0
,
BY ( 7 )
Proof.
thus
z c @ .
y 2 = lX,
Since
(7)
221
L,
a
1/2,
Corollary.
Let
12.31
1'11
>1;
Since
(1
then
1/2.
BY ( 4 )
= 1.
,
must either be. 1
(which is
or
la1
Assume that
2Re(r))
is in
a
E
L.
or
L.
By ( 5 )
-1.
Thus
Re(r)
T)
t
7 E L; is
proving the proposition.
T
is always in
T E
aD+.
aD+.
It will be convenient to consider
5 separate cases.
(1)
Case 1.
~ = u i , with
Case 2.
~ = i .
Case 3.
T=
Case 4.
T
Case 5 . Note that if which
-
eiel with
u > 1.
1r/3 < 8 < ~ / 2 .
+ i3&/2. ~ = p + v i = 1 / 2 +(34/2+v)i, =e
E p =
-r=i then
of course
-
L
1/2
with
v>O.
is the set of all Gaussian integers,
has many symmetries.
again has quite a few symmetries.
If
r = p
then
L
Cases 1,3, and 5 will be
called the general cases; cases 2 and 4 will be called the special cases.
12.32
By construction in (12.30)
222
Norman L. Alling Y
(1)
is dianalytically equivalent to
Let these two spaces be identified. lytic involution of @/LT
Let [6,
Let
(C/LT)/y.
be another diana-
y1
and let
be the field of all meromorphic "functions" on
El
p.12 ff.].
Y
Assume that
Yl
and
are dianalytically
equivalent; then there exists an IR-isomorphism h
E.
Since E(i) =F=El(i),
automorphism
h
of
F
F.
(yf
yr(f) = f
if and only if
is then-automorphism of
Thus, for all
El
h(i)=i: i.e.,
h
onto
is a
Since, by hypothesis, h
onto E l we have, for each
(3)
of
can be extended to an IR-linear
such that
@-linear automorphism of El
h
Y1
maps
fEF, y*(h(f)) =h(f). induced by
F
f € E l l yf(f) =f=h''y*h(f).
yl.) Since
y*(i) =-i= 1
h-ly*h (i), (4)
y f = h-'y*h.
Since h
is a C-linear automorphism of
analytic automorphism Riem@ (5)
6
of @/L,
F, there exists an
such that
h= 6*.
Since
is a contravariant functor (see e.g., [ 6 , p.99 ff.l),
Y1 = 6ys-I.
Conversely, given such an automorphism of @ / L T l Y
a dianalytic equivalence between anti-analytic automorphism p,
of
and @
Y1.
it engenders
y1
lifts to an
which must be of the
form (6)
where
p.,(z)
al E
= a l E + b1' @*
and
bl
for all
Z E @ ,
may be chosen to be in
.
P ( 1 , ~ ) (See
The Species and Geometric Moduli
223
(12.30) for more details.) (7)
p and y1 will be said to be equivalent if
Y
and
Y1
are dianalytically equivalent. Thus we have proved the following: and
Lemma.
exists
-6 ( z )
c
and
E @ *
cz + d,
take
dEC
for all
Assume that
are equivalent if and only if there
p
z
such that (5) holds: where E @ .
and
T1
are equivalent.
Since
onto another fundamental domain for
P(1,r)
L
6
must
in C ,
c
must satisfy the following. (8)
In case 1, c=+l. In case 2,
c=+ll
In case 3,
c = +-l .
In case 4,
c =+I,
In case 5,
c = +-l .
Note also (9)
or
+i.
or
+P,
or
2
f~
.
that in each case,
c/c=c 2 :
and that in the general cases (i.e., cases 1,3, and 51, 2 (10) c =l. Clearly 2 (11) 6y6 '(z) = c a z + c b + (d-ac2a) E Y , ( z ) = a 1z + b l , mod L, I _ _ -
for all
z EC.
It will be convenient to have established the following.
X - e i e l with (12)
z
(13) z
+ :A
=
8 EIR,
and let
-i8/2 2eiel2Re(ze 1I
z
E @ ;
Let
then
and
- X z = 2ieie/21rn(ze-i8/2)
Indeed, z
+ As =
-ie/2+zeie/2 i8/2-- (ze-i8/2+ (ze-ie/2) eie/2 (ze )e -
224
Norman L. A l l i n g
w,
Given a complex number
wt
w = 2 R e (w)
w
and
- w = 2 i im(w) :
p r o v i n g ( 1 2 ) and (13).
12.33
i s e q u i v a l e n t t o o n e and o n l y o n e of t h e
Theorem.
following:
1.1
Value of b -
Value of - T
Case ui,
with
I1
I1
I1
1.3
II
11
I1
11
I1
1.4
'I
1
2
-1
2
u > l
1.2
species
0
-1
0
2.1
i
1 ,L -1
2
2.2
If
i
-i
1
2.3
'I
1 'L -1
0
3.1
eiBl
3.2
I'
4.1
p :1 / 2 t i3'/2
with I1
5.1
p
1 , 3 , and 5 ) hence
a
v > 0
1
2 3
'Lp4
1
'LP5
1
1
1
-1
11
,
c2 = 1
(12:32:10); t h u s
is invariant.
(1)
i n cases 1 and 5 ,
(2)
i n case 3 ,
al=a
By (12.30:4 and 5 1 , ( a (= 1 ; a=tl;
hence and
a = +-. r .
Note a l s o t h a t (3)
if
1
0
Note t h a t i n t h e g e n e r a l c a s e s ( i . e . , i n c a s e s
By ( 1 2 . 3 0 : 7 ) ,
L.
T
P'LP
I1
Proof.
1
l'Lp
+viI with
5.2
-T
11
I1
4.2
in
n/3 < B < ?r/2
%
ac2 = 1,
then
d
- ac2a = 2 i I m ( d ) ,
(12.32:11) a
and
aT
,
and are
The Species and Geometric Moduli
(4)
if
ac2=-1,
then
225
d - a c 2d=2Re(d).
Having made these general observations, let us consider the several cases separately (in order of increasing complexity). Let
C a s e 1.
c = +-l .
(12.32:8), d
~ F u i ,with
- ac2d = 2iIm(d).
If
a=l,
Hence
d,
1 E L,
we may assume
and since
-
then
ac = 1 ;
is in
L;
1.1 and 1 . 3 . By (4),
Assume now that
d - ac2d = 2Re(d).
a=-1;
Hence
d,
our disposal, can be chosen so that is in L:
[O,ui). By (12.30:8),
i.e., either
values of
a
or
b=O
and
-
a
then
(12.30),
is in
is either b
0
in cases
acL=-l
(12.32:8).
which is completely at blr or equivalently b,
a6+b
(which is
b=ui/2.
b-6)
is in
This establishes the
in cases 1.2 and 1.4.
b
b
b
and
is real.
P(1,T)
that
hence
This establishes the values of
or 1/2.
By
thus by ( 3 1 ,
was chosen to be in
g+b
a = +-l .
blr or equivalently, b
without loss of generality
By (12.30:8),
[0,1).
b
By (11, L
which is completely at our dis-
posal, can be chosen so that Since
u > l .
Using ( 3 ) and (4),
one easily sees that cases 1.1, 1.2, 1 . 3 , and 1.4 are equivalent. As to the species [ui/2, 1+ui/2] s=2.
s
of
in case 1.1, clearly
[0,1] and
map down to two disjoint circles in
In case 1.2,
[O,ui] and
two disjoint circles in the species of
Y,
Y
period, and hence
aY;
[1/2, 1/2+uil
hence
s
is again
has no fixed points in
X.
thus
map down to 2.
in cases 1 . 3 and 1.4, note that y
aY;
Concerning b
is a half
Thus
s=O
in each case. C a s e 5.
By (12.32:8), hence
a
Let
T
c = +-l .
p+vi
(=1/2+ (3+/2+v)i),
By (l),
is an invariant of
y;
a = +-1 .
Since
with
c2=l,
v > 0.
al=a;
showing that cases 5.1 and
226
Norman L. A l l i n g
5.2 a r e i n e q u i v a l e n t .
Assume f i r s t t h a t
case 5.1 f i r s t ) .
By ( 3 ) ,
t i r e l y a t our d i s p o s a l , taken t o be real.
Since
b
thus
L;
is in
1
is either
b
c ( 35/ 4 + v / 2 ) i ;
-Z + C T . Let
p,
Thus
c=-1;
c=+1.
,
may
b,
i s en-
d
may be
(which i s
E+b
d
-
- further
be
2b)
b=1/2.
b e d e f i n e d t o be
= i + c / 2 + c ( 34 / 2 ) i + c v i = z + c ( p + v i ) =
T,
i s equivalent t o
hence we may t a k e
b
a s given i n c a s e 5.1.
t o be
showing t h a t i f
0;
i s e q u i v a l e n t t o a map covered by case 5.1. 2Now assume t h a t a = -1. By ( 4 ) , d - a c d = 2Re(d) ; t h u s b a=l,
then
be t a k e n t o be pure-imaginary.
Since
[O, ( 3 % +2 v ) i )
(since
t a k e n t o be i n
2 ~ - 1 = (4 3+2v)i.
By ( 1 2 . 3 0 : 8 ) ,
thus e i t h e r
or
b=0
is in
T
2.r
af;+b
b = (3$/2 + v ) i .
-1
b = ( 34 / 2 + v ) i .
A s noted above,
ac2 = -1.
=
p,(z)
d=-1/4;
b
may be
is i n
L
and
Let
c=l;
Since
then
pI ,(z)
d
d=-1/4;
then
T,(z)
a l e n t t o a map covered by c a s e 5.2. c a s e 5 . 1 i s indeed domain f o r (5)
Let
@/LT
1, in
C.
is
= -z - T ,
then
is entirely
=-;+T,
hence i t i s e q u i v a l e n t t o a map covered by c a s e 5.2. and l e t
7
then
L;
Assume t h a t
-; + ( 34/ Z + v ) i + 2 R e ( d ) .
a t o u r d i s p o s a l w e may l e t
L,
b = 0,
If
c=+1.
may
(=2iIm(b)) is in
e q u i v a l e n t t o a map covered by c a s e 5.2.
c=-1
is
Zero i s t h e v a l u e of
1/2.
A s s u m e , f o r a moment, t h a t
Let
p,(z)
then
b
L,
or
0
prescribed i n c a s e 5.1.
A s noted above,
Since
and hence e q u a l l y
By (12.30:8)
t a k e n t o b e i n [0,1). in
d-ac2z=2iIm(d).
bl,
( i . e . , consider
a = l
and
Now l e t
which i s equiv-
To s e e t h a t t h e s p e c i e s i n
let u s c o n s t r u c t a new fundamental Let
m - 3'+2v.
P l ( l , ~ ):{ x + i y : 0 < x < 1 and
max(-mx,mx-m) < y < m i n ( m x ,- m x + m ) I u
COI u
( 0 , ~ )u ( O , ~ - T ) .
The Species and Geometric Moduli
Clearly
is the set of fixed points of
[0,1)
-
z, and clearly
sidering the action of If
s=l.
Modify
on
y
C/LT
Y.
By con-
one easily verifies that
y.
then the imaginary axis is left fixed by
?=-z
to
P1(l,?)
P 1 ( l , ~ ) under
maps down to a circle in
[0,1)
-
227
so that
P2(1,.r)
[0,(3'+2v)i)
is in
P ( 1 , ~ )and ~ so that P2(l,~) is a rhombus that is a funda2 mental domain for @/L,. Then proceed as above to show that s is again
1. Let
Case 3 .
By (12.32:8) , showing that
T
:eiel
c = +-I ,
thus
a
T-
If
(7)
a = -T,
a=-.r; then
Since
and that
.
d - ac2i = 2eie/2Re(de-ie/21 .
then
= 0.
Re
As a consequence, al = a
One easily sees that
(2 + 2c0sB)'e~~/~, l=i(2-2cose)4eie/2
Assume that
gclR.
cL = 1.
+ 1=
T
71/3 < 8 < .rr/2. By (21, a=+T. -
is invariant, and thus that cases 3.1 and 3.2
are inequivalent. (6)
with
may be chosen so that
d
Having so chosen
T-1
is in
L,
d,
b = igei8/21
we may use (6) to show that,
without l o s s of generality, g
may be chosen to be in
[Ol(2-2cosB)4 ) .
By (12.30:8),
ac+b
(which is
this case) is in
L;
is in
L.
either
g=
g= 0
or
hence
2b
(2- 2cosB)'/2.
If
equivalent to a map covered by case 3.1. that
b f 0.
Let
by (12.32:11),
Now let ?,(z)
y1 ( z )
=
and let
-TZ + T ,
c = -1 and let
=-T~-T,
case 3.1.
c=l
d
5
for some
-Tg+b
in
As a consequence,
b = 0,
then
is
Assume, for a moment,
d - (2+2~ose)~e~'/~/4. , then, which is equivalent to case 3.1.
- (2 + 2~ose)'e~'/~/4- ,
then
which again is equivalent to a map covered by
Norman L. A l l i n g
228
(8)
If
a=T
Now assume t h a t thus
a=T.
g
r+l
is in
may b e c h o s e n t o be i n
[O, ( 2
r6 + b i s i n b = e i e / 2 ( 2 + 2cos9)'/2.
3.2. b
d-ac2a=2iei0/21m(de
may b e c h o s e n s o t h a t
d
(12.30:8), or
then
f o r some
b=gei9l2,
by ( 6 ) ; t h u s
2
c =1,
and
Assume t h a t
gt-IR.
It e q u a l s
L.
b = 0,
If
2b;
-i0/2
+2~0~9)').
)
=o;
By
hence e i t h e r
then
1.
and i s g i v e n
L,
b=0
f a l l s into case
t h e n w e see ( 6 ) t h a t
bfO;
Im(be
-i8/2
b= (~+1)/2.
t h e n i s a t t h e i n t e r s e c t i o n o f t h e d i a g o n a l s of t h e rhombus
whose v e r t i c e s a r e
and
0,1,~+1,
Let
T.
d r c i ( 2 - 2 ~ 0 ~ 9 ) ' e ~ ' / ~ (/ 4= c ( ~ - 1 ) / 4 ) ; t h e n by ( 1 2 . 3 2 : 1 1 )
,
( 6 ) , and ( 8 ) .
y
3 . 1 ( r e s p . c a s e 3.2), n o t e t h a t 0
r -1
and
=T;+CT;
i s e q u i v a l e n t t o a map
To see t h a t t h e s p e c i e s i s
c o v e r e d by c a s e 3 . 2 .
t h a t contains
y,
Hence
y,(z)
1
i n case
i s r e f l e c t i o n about t h e l i n e t h u s w e may p r o c e e d
.r+1);
(resp.
as w e d i d above f o r c a s e 5. Case 4 .
-
course then
in
G
Let
T ?
p ( - e in/3=1/2+i3'/2).
a p r i m i t i v e s i x t h r o o t of u n i t y .
i s a c y c l i c g r o u p of o r d e r 6 . By ( 1 2 . 3 0 : 7 and 4 ) ,
G.
is in
a
t h e n it i s a s u b g r o u p o f i n d e x phic t o t h e t w o element group be any e l e m e n t of d e t e r m i n e d by are
given
G,
a , only
2 2 4 G ( = { l , p ,P 1 )
mod G 2 and
Assume f i r s t t h a t
t o be
c L
al=c a
G
pG2
n
pn:
G/G2
Z);
E
c
is
2 2 ={g : g
E
GI;
i s isomor-
can be chosen t o (12.32:ll) i s
The two cosets of
i f it i s i n
of
G
a € G2
If
mod G
2
then
it i s a s cover-
C l e a r l y c a s e 4 . 1 and 4.2 a r e i n e q u i v a l e n t .
e d by case 4 . 2 .
a
Let
G.
Since
aE G,
:c
G
-
By ( 1 2 . 3 2 : 8 ) ,
and t h u s
Z2.
.
Let
2 pG ( = { p l p 3 , p 5 } ) .
i t i s a s ' c o v e r e d by case 4 . 1 ;
take
2,
is
p
1.
a
E
GL.
Since
By ( 1 2 . 3 0 : 8 )
c
i s a t o u r d i s p o s a l w e may
ag+b
is i n
L.
By c h o i c e
229
The S p e c i e s and Geometric Moduli
is in
b
is either
0
can choose
or b
i n case 4.1. t o be
a g + b = E + b = 2 R e ( b ) ; thus
P ( ~ , T ) (12.30).
y1
giving u s
Now assume t h a t
a
l o s s of g e n e r a l i t y .
Fix
2iein/61m(de-i'/6).
Since
g.
t h u s we may choose
g=O
If
but
L;
4 g=3/2.
s = l r let
y
Lastly let
a
5
1.
Since
-
G/G2
Since
(12.30:8) h o l d s , or
G.
then
or
b=1/2.
1E L ,
aE+b If
y
7
then
g = 3'/2.
yl(z)=pz+p,
i s complex con-
i s r e f l e c t i o n about
A s w e can see
p + l .
from t h e
1 i n e a c h case.
i.
G:{*l,+i}.
Let
By ( 1 2 . 3 2 : 8 ) , G 2 ( :cg2:
ac2 = 1 and so
C E
(which i s then
2b)
y
Clearly
- ac2a = d -
is i n
-
and
of
a l = %a. L e t
thus d
G
By
G.
g e G I ) ={+1}
w e may assume t h a t
b=O
a6+b
Assume t h a t
then
i s a t o u r d i s p o s a l w e may assume t h a t
(12.32:ll).
b=O
g= 0
I
34 e i v / 6 . ,
is
L,
iv/6
To see t h a t i n e a c h case
is T 5
ge
By (12.30:E)
i s t h e two element group;
c = +1;
Let
d
is i n
a
2ac d =
is
is i n case 4.2.
and
s
without
p,
-
d
b
i71/6 d=ie /4;
i s a c y c l i c group of o r d e r 4 .
course
+
[0,3 1.
a=p;
0
demonstration i n c a s e 5 ,
(12.30:4 and 7),
t o be
i n c a s e 4 . 1 and n o t e t h a t
let
may be chosen
By ( 8 )
Hence e i t h e r
andlet
the l i n e t h a t contains
Case 2.
1.
a
which i s i n
p+1,
and s o
I n case 4.2
jugation.
c2
Since
or equivalently
is i n class 4.2. a = l
may be chosen
i s e n t i r e l y a t our d i s p o s a l , w e
d
t o be i n
b=O
c-1
y
and hence
g
.
t o be
ag+ b=2b.
then Let
c
bl;
f o r some r e a l number
2
d
which i s e q u i v a l e n t t o
w e may choose
GLI
may choose it s o t h a t
pG
E
then
z + pr
then we a r e
b=O
If
1/2.
b=1/2;
(z) =
2d - a c d=2iIm(d), we
Since
or
0
Assume now t h a t
t o be any element i n
is in
c = l .
Let
t o be e i t h e r
i3'/4,
case 4 . 1 .
1/2.
Re(b)
b b
E
L;
= 2iIm(d).
i s real [0,1).
Since
hence e i t h e r
is i n case 2.1.
Since
230
c
Norman L. A l l i n g runs through
2 c =?l;
G,
a r e equivalent.
t h u s t h e cases
Assume now t h a t
i s as d e s c r i b e d by case 2.3.
a = l
Since
d
c a s e s 2 . 1 and 2.3 a r e i n e q u i v a l e n t .
c2 = il Let
and
a = + l and
i s pure imaginary,
a c{*iI.
Now assume
w e may assume, w i t h o u t loss of g e n e r a l i t y , t h a t
c 2 = 1;
t h e n by ( 8 )
may be c h o s e n t o be
'
( = 2 e i7T/4)
is i n
(12.30:8)
a6 + b
and h e n c e
g
i n c a s e 2.2. d E i2'eP'/4/4;
E
d - a c 2 z = 2 i e i T i l 4 I m ( d e- i ~ i / 4) ;
g e iTi/4, g
L,
L.
with
is either
0
or is g # 0.
Y1 ( z )
a 6 t b = 2b; 2'/2.
If
c = l
Let
2b
c l o s e indeed t o c a s e 1.3.
b
By E
L
7
is
and l e t
ig +i f and
which
To see t h a t
b=O
t h e a c t i o n i s t h e same a s c o v e r e d i n cases 1.1; t h u s Case 2 . 2 i s v e r y much l i k e c a s e 3.
thus
then
g=0
i s e q u i v a l e n t t o t h e map d e s c r i b e d i n cases 2 . 2 . a = l
a = i.
[0,2').
thus
(12.32: 11) e q u a l s
t h e s p e c i e s i s as c l a i m e d , n o t e t h a t i f
Since
1+i
may be c h o s e n t o be i n
Assume t h a t then
g E R . Since
In t h i s case
7
i.e.,
b=1/2;
- ac2J
b=O
then
s=2.
F i n a l l y c a s e 2.3 i s very
Thus t h e a r g u m e n t s above c a n b e u s e d
t o e s t a b l i s h t h a t t h e s p e c i e s a r e a s claimed i n c a s e 2 , proving t h e theorem. B i b l i o g r a p h i c note.
Most of t h e c o n t e n t s of t h i s c h a p t e r ,
up t o t h i s p o i n t , i s a n e x p a n s i o n o f
[6, pp.60-661.
W e have
s u p p l i e d a r g u m e n t s , m o d i f i e d n o t a t i o n , and made s l i g h t c h a n g e s . On p. 6 6 o f [ 6 ] t h e a u t h o r and Newcomb G r e e n l e a f made a n i n i t i a l s t a b a t g i v i n g a c o h e r e n t d e s c r i p t i o n of t h e moduli s p a c e s of r e a l e l l i p t i c curves
Y
whose c o n s t a n t f i e l d i s lR.
The i n i t i a l
a i m o f t h i s r e s e a r c h was t o g i v e a s i m p l e r p a r a m e t r i z a t i o n o f these spaces.
T h i s w e w i l l now do.
The Species and Geometric Moduli
231
12.34
(1)
Let
(2)
Let lR+s{tcn: t > O I .
For
denote complex conjugation.
K
a
in the upper half plane
T
(3)
let
XT :@/L~.
(4)
Let
Y
be a real elliptic curve whose field of constant
,
i.e.
is lR:
which has a non-empty boundary
aY
or which
is non-orientable. Let
Theorem.
exists a unique
5
t:t(Y)
3
~
cIR
.
Y of species 2 there
For each
+
such that
Y is dianalytically
equivalent to (5)
Xti/5 5
where
Y2,t
1
is the anti-analytic involution of
1
in case 2.1; also
in case 1.2; also
T=
induced by
in cases
T=i=T'.
i/t=-l/T'.
T o prove this theorem one need only look at the action
induced on
LT
in Theorem 12.33, for cases 1.1, 1.2, 2.1; and
shift to the action priate element in (6)
K
on
Lti, by multiplying by an appro-
@*.
A_ fundamental domain for
Let
ten+
0 - y - t/21.
and let
FD(Y
Y2,t. 2,t
)
-{x+iy: O O recall that
Y of species
For each
+ t-t(Y) cIR
such that
Y
there
1
is dianalytically
equivalent to
+' + ti/2/ 3%
(ii) t = 3 4 in case 4.1; also (34/2)i). (iii) 3 4 > t > l in case 3.1;
T=.r'.
( = + +
and
has species
m(X++ti/2)
ti/2.
f)z+
YlIt
=
r ; ( . r )I
where
0 .
(iii) g3
according as
t > 1, t = 1, o r t < 1.
(iv)
g2 > 0.
(v)
'p
maps t h e p e r i m e t e r o f t h e r e c t a n g l e whose v e r t i c e s a r e 0 , 1/2,
S p e c i e s , Geometric Moduli, Defining Equations
+
1/2
ti/2
and
ti/2,
injectively onto
255
(vi)
[-m,ml.
goes through t h i s s e t , counterclockwise s t a r t i n g a t
el
p(z)
0,
avoiding
+ ti/2)
but
> e3 ( i n ( t i / 2 ) ) .
)
e l , e 2 , and e 3
By Theorem 1 4 . 2 1 ,
Proof.
z
is s t r i c t l y decreasing; thus
p ( 1 / 2 ) ) > e2 ( :l p ( 1 / 2
(
0,
As
are real.
By
(7.33:4) t h e y are d i s t i n c t , proving ( i ) . Since 2
A = 16(e -e ) 1 2
( i i ) . For
i
-6
s
= -s
6
zero.
(e2-e,)
t = 1,
Recall t h a t
Hence
3
thus
g3(ti)
s
, A
(7.33:5)
6
and hence
s6, L
1 = 27g3/A
t > 1
> 0,
equals (7.33:18)
q6
(9.11:7)
,
proving
is
and t h a t
( C o r o l l a r y 5 , 99.28); t h u s
t > 1.
for a l l
t > 0,
-
(7.33:22)
1
i L = L;
J
for a l l
g3(ti) # 0,
(1) F o r
( e -e ) 2
showing t h a t
6'
J(ti) > 1
2
(i/t)Lti
= Lilt;
t = 1.
changes s i g n a t
Since
T
CD
I+
g3(LT)
i s a n a l y t i c (Theorem 9 . 2 2 1 ,
g3(ti)
shown t h a t
e x i s t s a n d i s p o s i t i v e [ 1 9 , p p . 6-
Limt++m
s6(Lti)
i s continuous.
7, a n d p p . 3 3 - 3 4 ] ; p r o v i n g ( i i i ) . S i n c e t > 1
(Corollary 5, 99.28))
for a l l
(3)
t > 1.
g2(Li,t)
proving ( i v ) .
,
and s i n c e
J(ti)> 1,
U s i n g (1) w e see t h a t 4
= t q2(Lti)
I
To see t h a t ( v ) a n d ( v i ) h o l d ,
Note i n p a s s i n g t h a t ( 2 ) and ( 3 ) i m p l y A(Li,t) 14.23
for a l l
3 J = g 2 /A, g 2 ( Lt l- 1 > 0 ,
F i g u r e 11, p r o v i n g t h e Theorem.
(4)
I t can be
= t
12
A(Lti)
Assume t h a t
,
for a l l
s = 1.
t > 0.
see [ 1 9 , p . 38
256
Norman L . A l l i n g
( i ) el
Theorem.
is real.
r e a l complex c o n j u g a t e s .
are distinct,non-
is positive,
( i i i ) g3
t > 1, t = 1, o r
t < 1.
(iv)
i s p o s i t i v e , zero, negative, zero, o r p o s i t i v e according
a s t > 3'12,
x
(v)
(O,l),
E
,
t = 3
> t > 0.
3-1/2, for
e3
and
2
(ii) A < 0 .
zero, o r negative according a s g2
e
t = 3- 1 / 2
3112 > t > 3- 1 1 2 ,
I
or b (x)
i s t h e r e l a t i v e minimum of
1/2
-
t h e r e l a t i v e minimum v a l u e b e i n g
of c o u r s e
-
e 1' el E ? ( 1 / 2 ) .
Proof.
el
By Theorem 1 4 . 2 1 ,
is real.
fi p ( 1 / 4 + t i / 4 ) . (Note: w e have 3 a d o p t e d t h e c o n v e n t i o n of numbering t h e e I s u s e d by [ 3 6 1 . )
e2
15 ( 3 / 4 + t i / 4 )
and
p
By Theorem 1 4 . 2 1 ,
e
i s r e a l - v a l u e d on
of r e f l e c t i o n about t h i s l i n e , a r e symmetric p o i n t s ; t h u s
3/4
+
-
e3 = e 2 ;
7Ri
ti/4
+
1/2.
and
I n terms
1/4
+
ti/4
p r o v i n g ( i ) . By ( 7 . 3 3 : 2 2 ) ,
2 2 2 A = 1 6 ( e -e 1 ( e 2 - e 3 ) (e3-el) , a n d by ( 7 . 3 3 : 4 ) t h e e I s a r e 1 2 j U s i n g (i) o n e e a s i l y sees t h a t A 5 0 , d i s t i n c t ; thus A # 0. proving ( i i ) . C l e a r l y L(
LT ,) ;
thus
(1)
( i / t ) L T I = L1/2
(2)
Hence
(3)
92 ( ~ 1 / 2t i / 2 t )
If
t = 1
'3 ( L 1 / 2
then
g3(L1/2+i/21
+ti/2)
equals
2 3 27g3 / ( 9 2
i/2tI
for a l l
= t 4 g 2 ('112
+
-
i/2
# 0.
< 1;
27g3)
t > 0.
6 = -t g 3 ( L 1 / 2 t t i / 2 )
+i/2t)
thus
,
+ti/2)
= T'.
As w e saw,
= 0.
J(1/2
+ti/2
+
1/2
t > 1,
93 (L1/2
is a basis of
{ti, 1/2-ti/21
I
and
*
( 2 ) t h e n shows t h a t
i n C o r o l l a r y 5 , 59.28, f o r J(LlI2
+
ti/2)
- 1,
0;
a n d so
i s less than
Using ( 2 ) w e see t h a t
which
t = 1 is t h e only
Species, Geometric Moduli, Defining Equations 93 (L1/2 + ti/2)
zero of
the rest of (iii).
DuVal [19, p. 33 ff.] establishes
Since
J(1/2+ti/2)
(Corollary 5, §9.28)), and since for all
t > 1.
Now let
T E Q
let
Since
< 1
for
t > 1 > 0
J = g2 3/A,g2(1/2+ti/2)
J ( p ) = 0 (9.28:7), g2(1/2+ 31i2i/2 = 0.
such that
1 ~ =1
1
and
be in case 3 of 512.3; then
T
257
0 < Re-r < 1/2:
(9.28).
(0,l)
J ( T )E
i.e.,
Using the Schwarz reflection principle, reflecting across the circle of radius J(1/2+ti/2)
1
and center
(0,lI
E
[19, p. 451.)
for all
1, we know that
31i2 > t > 3
(see e.9.
Using (3), we can establish the rest of (iv).
Using [19, pp. 38-391, (v) can be established, proving the Theorem. 14.24
Vorlesungen
Bibliographic note.
...
[64,
In Chapter 30 of Weierstrass's
pp. 264-2751 he applies some of his earlier
derived results and formulas to the case in which the numbers q2
and
q3
are real.
Many of the results presented thus far
in this Chpater can be found there.
For example, Weierstrass
notes that the study naturally breaks into two cases:
A
>
0,
el > e
and (11) A < 0. > e3.
In case (I) he noted that
is pure imaginary.
one of the
R
Further he noted that a basis
the period lattice could be chosen so that w2
e.'s 3
(I)
w1
5
(wp 2 )
t
of
is real and
In case (111, Weierstrass noted that
is real and that the others are a pair of
conjugate complex numbers. chosen so that it was
He further noted that Q could be (1 1/2 + ti/2) t , for some t > 0.
Most of the function theory theorems thus far presented in this chapter are well known.
See e.g., Chapter 2 and 3 of
DuVal's very useful little book [191.
Norman L . A l l i n g
258
- a s elsewhere i n P a r t
What may be n o v e l h e r e
-
monograph
I11 o f t h i s
i s t h e a s s o c i a t i o n of t h e a n a l y t i c function theory
o f r e a l e l l i p t i c f u n c t i o n s , a s found by E u l e r , L e g e n d r e , Gauss,
-
Abel, J a c o b i , W e i e r s t r a s s , Klein e t a l l w i t h t h e a l g e b r a i c
g e o m e t r i c t h e o r y o f K l e i n s u r f a c e s a s d e v e l o p e d by K l e i n , Witt, S c h i f f e r and S p e n c e r , A l l i n g and G r e e n l e a f , e t a l .
Species 0
14.30
Assume now t h a t
t > 0,
with
by u s i n g
9
and
s = 0; 5 K
9'
and
+
t h u s by ( 1 4 . 1 0 : l
1/2.
it i s
and 2 )
T I
5
ti,
Having m e t w i t h s u c h s u c c e s s
perhaps
a l i t t l e surprising to
learn that (1) 9
Indeed,
p
and
o ( ' p )( 2 )
a r e not i n 5
p e r i o d i c of period t r u e of
9I
,
( z +1 / 2 )
~'p
1/2,
E. = 'p ( z
+ 1/2).
w e see t h a t
'p
Since
o (lp) # 'p
.
is not
The same i s
e s t a b l i s h i n g (1).
Pedagogical
note.
I t seems t h a t o n e o f t h e g r e a t d i f f i -
c u l t i e s s t u d e n t s have when t h e y f i r s t b e g i n t o t r y t o do research i s t h a t mathematics, a s i t appears i n t e x t s , i n lect u r e s , and even i n r e s e a r c h j o u r n a l s , i s n o t o n l y p o l i s h e d ; b u t t h a t t h e i n v e n t o r s ( o r d i s c o v e r e r s ) o f t h e mathematics have been so t h o r o u g h a b o u t c o v e r i n g up t h e way i n which t h e i d e a s came t o them.
I n a n e f f o r t t o s h e d a l i t t l e l i g h t on how
some o f t h e r e s e a r c h was c o n d u c t e d i n a r r i v i n g a t t h e r e s u l t s o f t h e t h i r d p a r t o f t h i s monouraph, which may be of u s e t o s t u d e n t s and may a l s o b e of i n t e r e s t t o o t h e r s , 5914.31
-
14.33
a p p e a r s h e r e u s i n g t h e methods and t h e o r d e r o f t o p i c s a s t h e y a p p e a r e d i n t h e f i r s t d r a f t o f t h i s monouraph.
(Of c o u r s e
t h e y were much messier and more c o n f u s e d t h e r e , b u t t h i s i s
S p e c i e s , Geometric Moduli, Defining Equations
259
how t h e i d e a s e v o l v e d . )
One o f t h e s t a n d a r d a p p l i c a t i o n s o f t h e Riemann-
14.31
Roch Theorem i s t o show t h a t c e r t a i n e l e m e n t s e x i s t i n a n a l gebraic function f i e l d .
510.50 f o r a s t a t e m e n t of
(See e . g . ,
t h e Riemann-Roch Theorem i n t h e complex c a s e a n d e . g . , f o r t h e g e n e r a l t r e a t m e n t i n t h e complex c a s e .
[ 2 9 , 571
See e . g . ,
[ 1 6 , C h a p t e r 111 f o r t h e Riemann-Roch Theorem f o r g e n e r a l a l -
[4, 531 f o r t h e t h e o r e m i n
gebraic function f i e l d s , o r e.g.,
Our n o t a t i o n w i l l b e c o m p a t i b l e w i t h [ 4 ] . )
the r e a l case. yo
Let
be a p o i n t i n
y t i c Klein b o t t e ; thus f i e l d of c o n s t a n t s , field a t
yo
(1) L e t
b
then at
b yo
is not i n
yo
-
of
R,
is
E
i s IR- i s o m o r p h i c t o :-
i s of d e g r e e
ord b,
(3
The g e n u s
aY.
Even t h o u g h t h e
IR,the residue c l a s s (11.20).
C
X! y o 1 ;
i s a d i v i s o r on
(2
w h i c h w e know i s a d i a n a l -
Yo,t'
2
over
t h e o r d e r of g
b,
Yo,t
of
Since t h e residue c l a s s f i e l d
Yo,t'
IR, [ 4 , p.26 1 .
i s -2
i s , by d e f i n i t i o n , i (b)
To compute t h e i n d e x o f s p e c i a l t y
of
1.
b
(see e . g .
,
[4, p . 31]), w e may u s e t h e u s u a l d e v i c e , The S e r r e D u a l i t y Theorem (see e . g . ,
Yo,t (4)
such t h a t
[4,
3.91).
Given a d i f f e r e n t i a l
-
b > 0,
then
(w)
w
w
on
i s zero: t h u s
i ( b ) = 0.
By t h e Riemann-Roch
(5)
k(b) = 2:
(6)
i.e.,
L(b)
{f
Theorem (see e . g . ,
E
E(Y):
(f)
+
b
2
[ 4 , 3.81)
0)
, w e see
that
i s of d i m e n s i o n
260
Norman L . A l l i n g
2 over
IR.
Clearly
1 is i n
Clearly
{l,fj
element
h
and
b
(7)
h
7
poles a t
x
p
Let
-1
for
=
and
0
Clearly
and
xo
1,
with
and x1
Xti,
a
E
being
i s a map
h
A s a consequence,
( 1 0 . 4 0 ) o r [ 6 , p p . 95-1041,
[ C ( h ) : IR(h) ]
= 2
and
for
[F:E] = 2 ;
IR(h)] = 2 .
[E:
k
Let k
14.32
E
E - IR(h);
then
E = IR(h,k)
i s algebraic over
.
xo,
L e t us choose
and l e t
yo
i t s poles i n where
Thus w e h a v e
IR(h) o f d e g r e e 2 .
in
Yo,t
.
E (Yo, t)
t o be
let
pg(O),
q ( 1 / 2 ) = xl.
(1) W e want t o d e f i n e a n e l l i p t i c f u n c t i o n P(R) simple poles a t
R : (1 t i )t
0
Q
in
having
E
and 1 / 2 ,
.
I n 514.31 w e saw t h a t s u c h f u n c t i o n s e x i s t .
Fleierstrass zeta
f u n c t i o n s ( ( 7 . 3 0 : l ) and ( 7 . 3 4 ) ) a r e p a r t i c u l a r l y w e l l s u i t e d f o r t h i s p u r p o s e (Theorem 7 . 4 3 ) . (2)
Let
IR*
has its only
h
l e a r n e d s o m e t h i n g a b o u t d e f i n i n g e q u a t i o n s of
q(0)
b
( 1 2 . 3 6 ) and ( 1 3 . 1 5 ) ;
( ( 6 . 4 1 ) and ( 6 . 4 2 ) ) .
2
more d e t a i l s . )
Clearly
+
each being simple; t h u s
xl,
(See e . g . ,
(8)
{xo,x,}
(yo) =
j = 0
[F: C ( h ) 1 = 2 .
thus
Hence a n y o t h e r
IR.
af
L ( b ) - IR.
0'
of o r d e r
Xti
E
t h a t pole being a simple
Yo,t'
As a meromorphic f u n c t i o n o n
distinct.
of
y
F(Xti).
S(x.1
over
i s of t h e form
h a s o n l y o n e p o l e on
is i n
then
L ( b ) - IR
L(b)
f
Clearly
pole a t h
i s a b a s i s of
in
7R.
E
thus there exists
L(b);
Q ( z ) : i [ < ( z )- 5 ( 2
- 1/2)
- q1/2I,
for a l l
z
E
C,
S p e c i e s , Geometric M o d u l i , D e f i n i n o E q u a t i o n s
0.
Let
then
( S e e t h e e a r l y s e c t i o n s o f t h i s c h a p t e r f o r more
details.
'I
W'/W
= 1
136, p . 1901 (2)
and
w 5 1/4
(1) L e t
sn
is i n
+
w'
: ~ ' / 2= 1 / 4
h = -e -llt ,
ti,
sn(u)
E
IR, f o r a l l
m
1 + 2
u
E
u (sn) (u) : csn(u) = sn(u),
W e have s e e n (5.31:3) t h a t
=
then
a n d z = ei n v = e 2 r i u
L e t u s now c o n s i d e r L e g e n d r e ' s m o d u l u s
e3
ti/4;
iR.
E(Yl,t);
s i n c e (1) h o l d s ,
(3)
+
In=1 hn
2
k
f
2 2 R2/e3.
[36,p. 1961.
k.
establishing (2).
S p e c i e s , G e o m e t r i c Moduli, D e f i n i n g E q u a t i o n s
o3
(1),
h = -e -'t
Since
269
is real.
C l e a r l y t h i s can be w r i t t e n a s follows:
e2
(5)
= 2h
h = ei r
e- t r
thus
;
e-tr/4f
( 6 ) h1l4 = ei'/l (7)
k
E
2 hn -n
m
1/4
and h e n c e
lRi
proving t h e following:
k2 < 0.
k L = ( e 2- e 3 ) / ( e l - e 3 )
Since
are d i s t i n c t ( 7 . 3 3 : 4 )
, we
(14.34:3)
,
and s i n c e t h e
e 's j
have proved t h e following;
k2 < 0.
Lemma.
I t i s very easy t o prove t h a t
(8)
t h u s w e have
F(X . ) = C ( s n , s n ' ) ;
tl
E = IR(sn,sn'),
Theorem.
where
It is interesting t o notice t h a t t h i s differential
equation i s a s p e c i a l c a s e of Abel's d i f f e r e n t i a l equation. (10) where
2 2
c
e
and
14.43.
(1) c
2 2
( w ' ) ~= ( 1 - c w ) ( l + e w )
1
are
(4.12:3),
non-zero r e a l numbers.
Let
and
2 1/2
e : (-k )
;
t h e n A b e l ' s d i f f e r e n t i a l e q u a t i o n ( 4 . 1 2 : 3 ) becomes
Norman L . A l l i n g
270
( a s d e f i n e d i n ( 1 4 . 4 2 ) ) i s a meromorphic s o l u t i o n o f
sn (u
Abel s e l l i p t i c f u n c t i o n
r
radius
r :: m i n ( 1 , e )
Let
t o (2).
@
about
In
0.
(94.1)
i s another global solution
and l e t V
s i n g l e valued square r o o t .
be t h e open d i s c o f
V
t h e r i g h t hand s i d e o f ( 2 ) h a s a 2 2 2 1/2 be Let [ ( l - w ) ( l + ew ) I
t h e s q u a r e r o o t t h a t i s p o s i t i v e on
(0,r).
e q u a t i o n t h a t Abel c o n s i d e r e d on
is
$
s a t i s f i e s ( 3 ) [l, V o l .
sn'(u) = cn(u)dn(u)
[36, p. 2181; t h u s (3).
Since
w'
= 1/4
V
The d i f f e r e n t i a l
1, p . 2 6 8 1 .
[ 6 9 , p . 4921 and s n ' ( 0 ) = 1;
cn(0) = 1 = dn(0)
hence
sn
also satisfies
$ ( 0 ) = 0 = s n ( O ) , w e have p r o v e d t h e f o l l o w i n g :
r$ = s n ,
Theorem.
and
(2).
+
Corollary
ti/4,
where and where
E(YlIt)
Historical
note.
=
sn
is defined f o r
r$
w = 1/4
c = 1
i s defined f o r
n(@i@')
Since A b e l ' s paper i s w r i t t e n i n a
s t y l e t h a t i s n o t i n conformity with p r e s e n t standards of r i g o r i t c o u l d p e r h a p s b e a r g u e d t h a t Abel d i d n o t p r o v e t h a t h e had found a g l o b a l meromorphic s o l u t i o n o f (2).
After a l l
t h e n o t i o n o f a meromorphic f u n c t i o n h a s n o t b e e n f u l l y f o r m a l i z e d by 1 8 2 7 .
A t t h e very l e a s t it can be a s s e r t e d t h a t
sn,
as d e f i n e d h e r e w i t h t h e t a f u n c t i o n s , i s a s o l u t i o n o f ( 2 ) which e n j o y s a l l t h e p r o p e r t i e s t h a t Abel a s s e r t e d t h a t
$
had.
a u t h o r i s i n c l i n e d t o f e e l t h a t v i r t u a l l y e v e r y t h i n g on t h i s s u b j e c t a s s e r t e d by Abel and Gauss c a n b e p r o v e d w i t h o n l y a few a d d i t i o n a l comments.
A t t h e t i m e of p u b l i c a t i o n Abel's
The
S p e c i e s , Geometric Moduli, D e f i n i n g E q u a t i o n s
... was
Recherches
regarded,
271
for example by G a u s s , a s b e i n g
w r i t t e n a t a v e r y h i g h l e v e l of r i g o r .
(See O r e [ 5 2 ]
for
details. )
14.44
I n SVIII o f A b e l ' s R e c h e r c h e s
....
he t u r n s h i s
a t t e n t i o n t o t h e l e m n i s c a t e i n t e g r a l , 51.3, which Gauss h a d s t u d i e d e x t e n s i v e l y by 1 7 9 7 ( 4 . 3 1 ) .
lets
e = c = 1
function
9
To do t h i s A b e l m e r e l y
[l, V o l . I , p . 352 f f ] .
On d o i n g t h i s A b e l ' s
e q u a l s G a u s s ' s s i n l e m n (4.31:l).
This Page Intentionally Left Blank
CHAPTER 15
THE DIVISOR CLASS GROUP OF
Ys,t
Introduction
15.10
The divisor class group was considered for compact Riemann surfaces of genus
g
equals
in 558.61 and 8.62.
and
0
1
in 18.6, and computed explicitly when
g
Given a compact Klein
surface,one can define its divisor class group. 5.71 it was computed, using sheaf cohomology.
In [ 4 , 5.6 and
We will now make
an entirely independent computation of the divisor class group for
without the use of sheaf theory, which is both more
Ysltt
elementary and more explicit.
(1)
Let be
(2)
s = 2 , 1 , or 0, and let
Let
15.11 T '
= ti
if
1/2+ti/2,
Let
iL
K
is
s=2 if
or
0,
tclR,
with
t > 0.
and let it be defined to
s=l.
s = 2 or
1, and let it be
~ + 1 / 2 if
s=o.
Let
5
XT,
be the anti-analytic involution of
then
YsIt
(3)
5 {b: Y + Z such that s,t s,t on a finite subset of Yslt1.
Let
Let
is defined to be
XT,/(
(or E
for short)
b
is zero except
be the field of all meromorphic Given
let
associated with the point
Y
273
f
E
E*
and
YSlt,
"functions" on v
[6, 1.31. ys,t be the valuation of E
5;
(§12.3).
div Y
E(Yslt)
induced by
Y E
y.
274
Norman L. Alling
(Since y
is
contains IR, (4)
Let
(5)
then
u
by definition - a valuation ring of Y
group
E
F*-+ (f)
E*
that
is just its valuation.)
(f)(y) E uy(f),
f
E
for all
Y E
Ysrt;
is a homomorphism of the multiplication
into the (additive) group
div(Ysrt), whose
kernel is R*. A divisor in
(E*) is called a principal divisor.
The most fundamental question is to find necessary and sufficient conditions for
b
div Y
in
s,t
to be principal.
A
closely related question is to compute (6)
C(Ysrt) div Y srt/(E(Ysrt)*)r
To do this it will be useful Ys,t. to define the notion of the degree of b. Roughly this measures the divisor class group of
the number of zeros of
b
minus the number of poles of
However, the degree of
b
must also take into account the degree
of the residue class field of each [16, 1.71 for details.)
is either IR
where
2 cy
E
(See e.g.,
Since the residue class field of
or is IR-isomorphic to
correct definition of the degree of (7)
over IR.
y
b.
supp (b) - ay b(y) + ‘y
supp(b), the support
of
E
cr
y
the following is the
b:
supp (b) n ay b(Y) r
b,
is
{ Y E Y: b(y) f 0 1 ,
and
we understand that the sum over the empty set in (7) is the integer zero. onto
2.
(8)
Let
Clearly
deg
is a homomorphism of
div Y
srt
divO Ysrt: ker deg.
A divisor in shortly that
div0Ysrt is called homogeneous.
We will see
The Divisor Class Group of
(91
275
YsIt
(E(YsIt)*)cdivoYsIt. divoYs,t/(E(Ys,t)*) z
(10) Let
c 0 (Ys,t)'
This will be called the homogeneous divisor class group of ys,t. C(Ys,t)/Co(Yss,t) = Z.
(11) Clearly
In [ 4 , 5 . 7 1 it was shown that (12)
CO(YSIt) is isomorphic to
@/Z)
that it is isomorphic to I R / Z
Calculations gg
15.20
~EJZ 2
if
s= 2,
if
s=l
or
and
0.
xTl
Having made definitions and stated results (15.11:9 and 1 2 ) directly on
YsIt, let us proceed up to the covering space
X T , to prove them. and
p
c*(a) 5 a5 5*
(2)
is again in
div X T l
is an involution of
involution of
YsIt.
XT1
(See (13.15) for de-
a € div X T I ,
Given
Clearly
XT1
is an anti-analytic involution of
is its quotient map onto
tails.)
(1)
5
divO X T I
.
.
div X T ,
which induces an
(See §8.6 for details on divisors on
-1
X T l :{a E div E X i 1 : c*(a) = a } 5 sym divoXTl:sym div X T , n divoXTl 5 5
Let
sym div
and let
.
Clearly each of these sets is an additive group.
X T I onto (3)
and
p*(b) p*
Ysrtl
= bp
given is in
b
E
Since
div Y srt
div X T l ,
is an injective homomorphism.
Lemma.
(i) p*
maps
div Y srt
injectively onto
p
maps
Norman L. Alling
276
.
(ii) For all b sym div XT , 5 (iii) p* is an injection of
is injective. I
/5
5
Since
Ys , ,
div Ysft
into
a=p*(b); let
is in
aY;
point
x
a=bp.
then
p* (b),
Y E
Let
b
P*
X T l onto p*
proving that
maps
j= 0
3
deg a = 2.
Y
b c div Ysft; thus and, for the moment,
s,t
is in
X T l . Clearly
S(x.1 = x ~ - ~ for ,
p*(b)
=
xfxl
5
a,
and thus
then (15.11:7) deg b = 2,
j and
is
in
1.
X T l , xO#xl,
Thus
deg a = l .
and
but
p*(b) = x I x o 3 + x f x l l!a ,
Since elements of the form
div Ys,t,
basis of
sym divOX 5 is surjective,
div Ys,t. Assume first that y -1 deg b = 1 and p (y) consists of a single
then
in
=
Clearly
Assume now that y E Y - aY; -1 p (y) = Cxo,xll, where x
and
onto
is the quotient map of
proving (i).
x{YP
b:
deg b = d e g p*
sym div X T , . Let a E sym div X T l ; then 5 5 a = a 5 ; thus a induces a map b of XT,/C
such that
2,
p
S*p* (b) = bpg = bp
a : X T , + Z and into
,
div Ys,
divO ys,t We have noted that since p
Proof.
xT
E
(ii) is proved.
form a free XIYl (iii) then follows,
proving the Lemma.
E(
15.21
3
E(Y
may
))
t F( 5 F(XT , ) ) . Sf
subfield of
(1) o(f)
to be
K
f 5,
having
E
(F*) c divOXT (2)
(E*)
c
Lemma. Proof.
of course - be regarded as a
In fact, if we define for each
as we did in 514.10, then F
-
a
f
E
F,
is an IR-linear automorphism of
as its fixed field.
By Theorem 6.41,
(as noted again in (8.62:l) )
divoXTI
;
thus
.
(E*) = (F*) n sym divOXT,. 5 Let g E E* and let a: (g).
Since
g
E
F*, a
is
The Divisor Class Group of in
Since
(F*).
E* c (F*) n sym divOXTI . 5 (3)
let
f
be in
Let
u(f)/f - h ; then
such that
F*
h
=
Ehf;
hence
Let
1 hl
h = ei8
such that (4)
h
=
1.
and
Clearly
is in
sym divoX,,.
5
(h)=5*(d) - d = O .
By
is in C * .
Clearly a(f) =hf. 2 f = u (f)= u(hf) = g u ( f )
F,
Thus there exists a unique
then
g
E
E*
and
Indeed, u (9)= e-ie/20(f)=eiel2f = g ,
E*.
(15.20:l); and so
8,
02
6
0.
m,
are in
j
as
If
a=m.
n=4
a'
aY.
If
n=3
then
P(a)
then
P(a')
r/2
0,
e1=2
then and
has the same sign
ranges through IR.
Thus there exist exactly
0
then
s'=r'/2.
n
then
aY.
Assume that
is either positive definite or negative
definite; accordingly
s'=2
or is
0.
The proof is essentially that given in 517.20. Y'
T h e o r e m (Alling-Greenleaf)
if
r' = n ,
or
Let
Proof.
on which
Cm}
and if
Q(x)
over IR(x)
Y'
is positive definite. Y'
then
can be obtained by gluing
together along the
Q(x)'O;
hence
occurs over
Y'
Y'
intervals in If
r' = O
Q:
-IR.
Since
consequence Y '
E'
is orien-
Riemc E' (i) :X I
can be regarded as a closed subspace of
0 < rl
occurs over C -IR
'
[ 3 6 , p.2181.
[ 3 6 , p.2281;
thus
proving t h a t (14)
A(T)
has a simple zero a t i n f i n i t y .
Using t h i s , t h e f a c t t h a t the fact that for a l l A
m c
A
r2r
i s a n a l y t i c on X (m(-r)) = A
w i l l b e c a l l e d a u t o m o r p h i c 0" 9)
with
(T)
4
( 1 7 . 4 2 : 2 ) , and
(Theorem 1 7 . 4 2 )
respect
to r 2 .
(A
,
From D e f i n i n g E q u a t i o n t o S p e c i e s and Moduli
Q
i s sometimes c a l l e d a modular f u n c t i o n on
r2.)
a subgroup Since
(See e . g . ,
transcendental extension of
@(J)
i s a s u b f i e l d of
(15)
@ ( j )c C(T)
b u t o n l y under
[251 f o r d e t a i l s . )
c(A)
i s n o t a c o n s t a n t (13),
A(T)
even though it i s
r,
n o t i n v a r i a n t u n d e r t h e f u l l modular g r o u p
313
i s a pure
By ( 9 ) o r ( 1 0 ) w e see t h a t
@.
and t h a t
@(A)
i s an a l g e b r a i c e x t e n s i o n of d e g r e e a t most
6.
J = a2 + 1
I n t h i s c o n t e x t ( 8 ) and t h e f a c t t h a t
c a n be g i v e n
t h e following i n t e r p r e t a t i o n :
(16)
C(J)
Since
a2
c
@(a)
c
+ 1- J = 0 ,
2 o r 1.
@(A).
c ( a ) over
t h e d e g r e e of
By Theorem 17.41
(vi),
a ( . r + l ) =-a(?);
r;
i n v a r i a n t u n d e r t h e a c t i o n of
hence
is either
@(J) thus
a
is not
As a c o n s e -
a k @(J).
quence [@(a): @(J) I = 2 .
(17)
r
W e have s e e n t h a t t h e modular g r o u p i n g t o (l), and t h a t t h e s u b g r o u p o f
is let
r2
(Theorem 17.42).
f(A)
(18)
Let
thus
r
(19)
Let
Let
m c
r
be a r a t i o n a l f u n c t i o n i n
r
t h a t leaves
and l e t A
a c t s on
f(A)
E
A
accord-
A
fixed
@(A):
i.e.,
w i t h complex c o e f f i c i e n t s
m(f ( A ) ) :f ( m ( A ) ;
acts a s a group of G
C-automorphisms o f
d e n o t e t h e g r o u p o f c-automorphisms o f
i n d u c e d by t h e a c t i o n o f isomorphic t o
r2;
r;
thus
G
c(A)
is naturally
r/r2.
I n 517.42, w e saw t h a t having k e r n e l
@(A).
thus
T
i s a homomorphism o f
r
onto
S3
Norman L. A l l i n g
314
G
(20)
and
on
a r e n a t u r a l l y i s o m o r p h i c , t h e a c t i o n of
S3
b e i n g g i v e n by ( 1 ) . L e t
C(A)
G
and
s3
be
S3
identified. Let
F
be t h e f i x e d f i e l d of
t o conclude t h a t [@(A):
(21)
C(J)] C(J) =
5
~ ( a )is
Indeed, s i n c e
But w e saw i n ( 1 9 ) t h a t
thus [ C ( A ) : C ( J ) l = 6.
and
By c o n s t r u c t i o n ( 4 ) (22)
G(T)
i s f i x e d under
t h e f i x e d f i e l d of a=-3-3/2Gl
a
A3.
A3.
is i n the fixed f i e l d
Using a l i t t l e G a l o i s t h e o r y w e know t h a t a l s o know t h a t
C(a) c
[c(A): C ( J ) ] = 6 Since
A,
that
( 2 1 ) ; hence
C ( a ) [XI
d u c i b l e polynomial i n
-
2x3
Thus
C(X)
( 3 - 33/2a1x2
[ C ( A ) : A ] = 3.
[ C ( a ) : @ ( J )=]2
C(a)
A,
equals
(17),
A3.
W e
and t h a t
establishing (22).
which
i s a r o o t (8):
X
- ( 3 + 33/2a)x + 2;
i s t h e s p l i t t i n g f i e l d of (23) over
The subgroup o f
of
A
(8) g i v e s r i s e t o t h e following irre-
[ C ( A ) : @ ( a ) ]= 6 ,
(23)
@ ( J c) F .
i s 6 w e may u s e s t a n d a r d G a l o i s Theory
S3
[ C ( h ) : F] = 6 .
6;
F
J(.r)-1=a2(-c) =
a s w e have s e e n i n t h i s s e c t i o n ,
G L ( - c ) / 2 7 = -F(-c)/27,
Since t h e o r d e r of
Since
S3.
r
@(a).
t h a t induces t h e a c t i o n
on
A3
@(A)
is (24)
T - ~ ( A ~E )
r;/r3
a r2.
=A ~ .
r2 r;
r:
i s a s u b g r o u p of i s a subgroup of
r
o f i n d e x 3 and o f i n d e x 2.
Thus w e have p r o v e d t h e f o l l o w i n g r Theorem.
a
i s a u t o m o r p h i c on &
I n Example 3 o f 117.42, w e saw t h a t a(q),
e q u a l s (1 3 2 ) :
hence
a(q)
with respect t o
n(Q),
crenerates
ri.
and t h u s A3.
A s a con-
From D e f i n i n g E q u a t i o n t o S p e c i e s and Moduli
315
sequence
r2
(25)
and
r;
Since
q
g
r,
i s o f i n d e x two i n
g i v e n any
(26)
r;.
generate
r
g c
- r;,
rz
t h e g r o u p g e n e r a t e d by
and
r.
is
F i n a l l y w e have: for all
(27)
g
E
r
and a l l
T E
4,
a(
= +a (
g ( T ) )
T )
,
the sign
~ ( g )i s a n e v e n p e r m u t a t i o n i n
being p l u s i f
minus i f i t i s a n odd p e r m u t a t i o n i n
and
S3
s3'
I n d e e d , t h i s f o l l o w s from ( 1 7 . 4 1 : 4 ) .
17.44
v a l u e of
t
Let
We w i l l now c o n c e r n o u r s e l v e s w i t h t h e
> 0.
a ( t i ) and
a (1/2
The map
Theorem 1.
t EIR'
properties:
(i) it h a s r a n g e IR,
increasing,
( i i i l it i s a
(v) a ( i / t ) = - a ( t i ) , Proof.
g3(ti)
.
+ ti/2)
a (ti) has t h e following
+
( i i ) it i s s t r i c t l y monotone
C(")-map,
for a l l
( i v ) it t a k e s 1 t o 0 , and
+.
t EIR
& ( t i> )0.
By Lemma 1 of § 1 7 . 4 0
By Lemma 1 7 . 4 1 ,
is p o s i t i v e , zero, o r negative according as
or
t < l ; thus
as
t >1, t = l , o r
a(ti)
t
0,
:2ai,
T
e3
and
k c (0,l).
let
and let
K ( T )
2
:e2/e3
2
being defined in terms of
T
((5.20:2,3 and 4) and
0
such that
.
(5.30:3) )
Lemma.
There exists a unique
Proof.
It is easier to consider
modulus to Legendre modulus
K
a
>
k =K(T).
the complementary
K ' ( ; ) ~
(See e.y. , (1.10:3) and
(;).
(3.31:6) for other references to this modulus.) (4)
K~
+ ( K ' )= ~1 and
K '
= Oo/e3 2 2
Using the product development of
@
[36, p.2141. and
0
e3
[36, p.2041 we
find that
Since K '
(G)
ii is a function of a, namely e-2na., thus is a function, F(a) 4 , of a. Clearly it is continuous =2ai,
and differentiable on IR+ always positive.
.
Since
a
>
0, 0 < h
0, the geometric modulus of Ys,t' Chapter 1 2 , 5 1 2 . 3 t E h(T), 5 9 . 2 3 T E
W'/W
5 W2/W1,
§5.20:1
ti if s = 2 or 0, and 1 / 2 + ti/2 if Theta functions, Chapter 5 and 3, 5 5 . 2 0 : 4 ej(v) , for j = O , 1 , 2 , e;,e2,e3, and e4, § 5 . 3 0 : 3 Trace, T,(f), of f, § § 7 . 4 2 : 5 and 11.11 Tr(f), the trace of f, § § 7 . 4 2 : 5 and 11.11 T'
f
s = l , §13.16:1
Index
349
Transition functions, §8.20:4 Trans @ , the translation subgroup of @ , §13.11:3 Translation by b , Tb : , §§3.13:6 and 17.32 Translation of c by b, gb, 513.11:2 Transpose Mt of a matrix M Triangular, extreme, and median lattices, 517.52:6
(i y )
u
('1 y)
,
and u h(U), §9.23:7 U(A) , the group of units of the ring A, 53.11 U(0) (or U for short), the group of units of the valuation ring 0, §10.12:1 U §13.20:1 Y' Upper half plane, Q, §3.12:4 f
Vx, §13.30:1 Value group GO (or G for short) of 0, §10.11:5 Valuation ring, 510.11:5 Valuation ring at m, om, 510.13, Example 2 (cont.) Vertical lattices, §17.52:5 W ( s ) f 4s3 - g2s - g3, Weierstrass form, §7.20:2 2' "3 WL(X) f 4x3 - g2 (L)x - g 3 (L)I 517.51:l Weierstrass's Addition Theorem, 58.52:20 Weierstrass's inversion problem, 99.50 Weierstrasslsl)-function, 997.21 and 7.32:4 Weierstrass's o-function, §§7.21 and 7.32:2 Weierstrass's t-function, §§7.30:1 and 7.34:l (X,p,S), the complex double of Y, 513.15 6 , 913.15 5 : K if s = 2 or 1 and K + 1/2 if s = O , §13.16:2
-
Yo,t' 512.36: 1 Yl,t' 512.35 :1 512.34:5 Y2,t' Z, the ring of integers or its addition group